Integrals: Overview, Questions, Preparation

• Integration as an Inverse Process of Differentiation
• Methods of Integration 
• Integrals of Some Particular Functions 
• Integration by Partial Fractions
• Integration by Parts 
• Definite Integral 
• Fundamental Theorem of Calculus
• Evaluation of Definite Integrals by Substitution 
• Some Properties of Definite Integrals

Check the NCERT Solutions Maths Class 12 Chapter 7 Integrals below.

Download Here: NCERT Solution for Class XII Maths Integrals PDF

Ex.7.1

Find an anti-derivative (or integral) of the following functions by the method of inspection.

Q1.  $sin2x$

A.1.

$\begin{array}{l}\frac{d}{dx}cos2x=–2sin2x \\ sin2x=–\frac{1}{2}\frac{d}{dx}\left(cos2x\right) \\ sin2x =\frac{d}{dx}\left(\frac{1}{2}cos2x\right)\end{array}$

Therefore, an anti-derivative of  $sin2x is-\frac{1}{2}cos2x$

Q2.  $cos3x$

A.2.

$\begin{array}{l}\frac{d}{dx}sin3x=3cos3x \\ cos3x=\frac{1}{3}dx\left(sin3x\right) \\ cos3x= \frac{d}{dx}\left(\frac{1}{3}sin3x\right)\end{array}$

Therefore, an anti-derivative of  $cos3xis\frac{1}{3}sin3x$

Q3.  $e^{2x}$

A.3.

$\begin{array}{l}\frac{d}{dx}\left(e^{2}x\right)=2e^{2x}\\ e^{2x}=\frac{1}{2}-\frac{d}{dx}\left(e^{2x}\right)\\ e^{2x}=\frac{d}{dx}\left(\frac{1}{2}{e}^{2x}\right)\end{array}$

Therefore, an anti-derivative of  $e^{2x}is\frac{1}{2}{e}^{2x}$

Q4.  ${\left(ax+b\right)}^2$

A.4.

$\frac{d}{dx}{(ax+b)}^3=3a{(ax+b)}^2$

${\left(ax+b\right)}^2=\frac{1}{3a}.\frac{a}{dx}{(ax+b)}^3$

${\left(ax+b\right)}^2=\frac{d}{dx}\left(\frac{1}{3a}{(ax+b)}^3\right)$

Therefore, an anti-derivative of  ${(ax+b)}^{3}is\frac{1}{3a}(ax+b){}^3$ .

Q5.  $sin2x–4e3x$

A.5.

$\frac{d}{dx}\left(-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}\right)=sin2x-4e^{3x}$

Therefore, an anti-derivative of  $\left(sin2x-4e^{3x}\right)is-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}$

Find the following integrals in Exercises 6 to 20:

Q6.  $\int \left(4e^{3x}+1\right)dx$

A.6.

=  $\begin{array}{l} {\displaystyle \int 4e^{3x}dx}+{\displaystyle \int 1dx}\\ \frac{4·e^{3x}}{3}+x+C\end{array}$

Q7.  $\int x^2\left(1-\frac{1}{x^2}\right)dx$

A.7.

$\begin{array}{l}\int x^2-\frac{x^2}{x^2}dx\\ ={\displaystyle \int x^2}–1dx={\displaystyle \int x^2}dx–{\displaystyle \int 1dx}\end{array}$

$=\frac{x^3}{3}-x+C$

Q8.  ${\displaystyle \int \left(a{x}^2+bx+c\right)dx}$

A.8.

$\begin{array}{l}=a.{\displaystyle \int x^2.}dx+b{\displaystyle \int x}.dx+c{\displaystyle \int 1.dx}\\ =\frac{a·x^3}{3}+b\frac{x^2}{2}+cx+d.\end{array}$

Q9.  ${\displaystyle \int \left(2x^2+e^x\right)}dx$

A.9.

${\displaystyle \int 2x^2+e^x}dx$

$\begin{array}{l}=2·{\displaystyle \int x^2}dx+{\displaystyle \int e^x}dx\\ =\frac{2·x^3}{3}+e^x+\; C\end{array}$

Q11.  $\int \frac{x^3+5x^2-4}{x^2}dx$

A.11.

$\begin{array}{l}\int \left(\frac{x^3}{x^2}+\frac{5x^2}{x^2}-\frac{4}{x^2}\right)dx\\ \begin{array}{ll}={\displaystyle \int (x+5-4x^{–2})}dx\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int x}.dx+{\displaystyle \int 5}dx–{\displaystyle \int 4x^{–2}}dx\hfill \end{array}\\ =\frac{x^2}{2}+5x-4\frac{x^{-2+1}}{-2+1}\\ =\frac{x^2}{2}+5x-\frac{4x^{-1}}{-1}\\ =\frac{x^2}{2}+5x+\frac{4}{x}+\; C\end{array}$

A.12.

$\begin{array}{l}={\displaystyle \int \left(x^3.x^{–1/2}+3x.x^{–1/2}+4.x^{–1/2}\right)}dx\\ =\int \left(x^{5/2}+3x^{1/2}+4x^{-1/2}\right)dx\\ =\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+3·\frac{x^{1/2+1}}{\frac{1}{2}+1}+4·\frac{x^{-1/2+1}}{-\frac{1}{2}+1}\\ =\frac{x^{7/2}}{7/2}+\frac{3·x^{3/2}}{3/2}+\frac{4x^{1/2}}{1/2}+C\\ =\frac{2}{7}{x}^{7/2}+2x^{3/2}+8x^{1/2}+C\\ =\frac{2}{7}{x}^{7/2}+2x^{3/2}+\mathrm{8\surd x}+C.\end{array}$

Q13.  $\int \frac{x^3-x^2+x-1}{x-1}dx$

A.13.

$\begin{array}{l}\int \frac{x^3-x^2}{x-1}+\frac{x-1}{x-1}dx\\ \int \left(x^2\frac{(x-1)}{x-1}+1\right)dx\\ \begin{array}{ll}={\displaystyle \int \left(x^2+1\right)}dx\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int x^2}.dx+{\displaystyle \int 1}.dx\hfill \end{array}\\ =\frac{x^3}{3}+x+\; C\end{array}$

Q14.  ${\displaystyle \int (}1-x)\surd xdx$

A.14.

$\begin{array}{l}\int (\surd x-x·x^{1/2})dx\\ \int (\surd x-x^{3/2})·dx\\ \int x^{1/2}·dx-\int x^{3/2}·dx\\ \frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}+C\\ \frac{2}{3}{x}^{3/2}-\frac{2}{5}{x}^{5/2}+\end{array}$ c

Q15.  $\int \surd x\left(3x^2+2x+3\right)dx$

A.15.

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^2.x^{1/2}+2x.x^{1/2}+3x^{1/2}\right)}.dz\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^{5/2}+2x^{3/2}+3x^{4/2}\right)}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 3x^{5/2}}.dx+{\displaystyle \int 2x^{3/2}}.dx+{\displaystyle \int 3x^{1/2}}.dx\hfill \end{array}\\ =3·\frac{x^{7/2}}{7/2}+\frac{2·x^{5/2}}{5/2}+\frac{3·x^{3/2}}{3/2}+\; C\\ =\frac{6·x^{7/2}}{7}+\frac{4x^{5/2}}{5}+\frac{6}{3}{x}^{3/2}+\; C\\ =\frac{6x^{7/2}}{7}+\frac{4·x^{5/2}}{5}+2x^{3/2}+\; C.\end{array}$

Q.16.  ${\displaystyle \int \left(2x–3cosx+e^x\right)}dx$

A.16.  $={\displaystyle \int 2x}dx–{\displaystyle \int 3cosx}+e^{x}dx$

$\begin{array}{l}=\frac{2x^2}{2}-3sinx+e^x+\; C\\ =x^2-3sinx+e^x+C\end{array}$

Q.17.  ${\displaystyle \int (2x^2–3sinx+\mathrm{5\surd x})}dx$

A.17.

$\begin{array}{l}={\displaystyle \int 2x^2}.dx–{\displaystyle \int 3sinx}.dx+{\displaystyle \int \mathrm{5\surd x}}dx\\ =\frac{2x^3}{3}+3cosx+\frac{5x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C\\ =\frac{2x^3}{3}+3cosx+\frac{10x^{3/2}}{3}+\; C\end{array}$

Q18.  ${\displaystyle \int secx\left(secx+tanx\right)}dx$

A.18.

$\begin{array}{}\end{array}={\displaystyle \int \left(se{c}^{2}x+secxtanx\right)dx}$

$={\displaystyle \int se{c}^{2}x.dx}+{\displaystyle \int secxtanxdx}$

$=tanx+secx+C$

Q19.  $\int \frac{se{c}^{2}x}{cose{c}^{2}x}dx$

A.19.

$\begin{array}{l}\int \frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}dx\\ =\int \frac{si{n}^{2}x}{co{s}^{2}x}dx\\ \begin{array}{l}={{\displaystyle \int tan}}^{2}xdx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(se{c}^{2}x–1\right)}dx={\displaystyle \int se{c}^2}xdx–{\displaystyle \int 1}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=tanx–x+C\hfill \end{array}\end{array}$

Q20.  $\int \frac{2-3sinx}{co{s}^{2}x}dx$

A.20.

$\begin{array}{l}\int \left(\frac{2}{co{s}^{2}x}-\frac{3sinx}{co{s}^{2}x}\right)dx\\ \begin{array}{l}={\displaystyle \int \left(2se{c}^{2}x–3tanx.secx\right)dx}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 2se{c}^{2}x}.dx–{\displaystyle \int 3tan}x.secx.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=2tanx– 3 secx+C\hfill \end{array}\end{array}$

Choose the correct answer in Exercises 21 and 22.

A.21.

$\frac{x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{x^{1/2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C$

$=\frac{2}{3}{x}^{3/2}+2x^{1/2}+$ ⸫ The correct Answer is (C)

Q.22.  $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4},f(2)=0$

A.22.

$\begin{array}{l}f\left(x\right)={\displaystyle \int 4x^3}–\frac{3}{x^4}dx\\ =4\cdot \frac{x^4}{4}-3\cdot \frac{x^{-3}}{-3}+\; C\\ =x^4+\frac{1}{x^3}+\; C\end{array}$

Now,  $f\left(2\right)=0 \therefore f\left(2\right) =2^4+\frac{1}{2^3}+C=0$

$\begin{array}{l}=16+\frac{1}{8}+c=0\\ =c=-\left(16+\frac{1}{8}\right)\\ =c=-\frac{129}{8}\end{array}$

Therefore, correct answer is A.

Ex.7.2

Integrate the functions in Exercises 1 to 37:

Q1.  $\frac{2x}{1+x^2}$

1.

$\begin{array}{l}Let,1+x^2=t\\ \Rightarrow 2xdx=dt\\ I=\int \frac{2x}{1+x^2}dx=\int \frac{1dt}{t}\\ \Rightarrow log\left|t\right|+c\\ \Rightarrow log\left|1+x^2\right|+c\end{array}$

Q2.  $\frac{{(logx)}^2}{x}$

A.2.

$\begin{array}{l}Let,logx=t\\ \frac{1}{x}dx=dt\\ I=\int \frac{{(logx)}^2}{x}dx=\int t^{2}dt=\frac{t^3}{3}+C\\ =\frac{{(logx)}^3}{3}+C\end{array}$

Q3.  $\frac{1}{x+xlogx}$

A.3.

$\begin{array}{l}\frac{1}{x(1+logx)}\\ Put,1+logx=t\\ =\frac{1dx}{x}=dt\\ =I=\int \frac{1}{x+xlogx}dx=\int \frac{1}{t}dt\\ =log\left(t\right)+c\\ =log\left(1+logx\right)+c\end{array}$

Q4.  $sinxsin\left(cosx\right)$

A.4.

Q5.  $sin\left(ax+b\right)cos\left(ax+b\right)$

A.5.

$\begin{array}{l}=sin\left(ax+b\right)cos\left(ax+b\right)=\frac{2sin\left(ax+b\right)cos\left(ax+b\right)}{2}\\ =\frac{sin2(ax+b)}{2}\\ Put2\left(ax+b\right)=t\\ 2adx=dt\\ I=\int \frac{sin2(ax+b)}{2}dx\\ =\frac{1}{2}\int \frac{sintdt}{2a}=\frac{1}{4a}[-cost]+C\\ =-\frac{1}{4a}cos2(ax+b)+C\end{array}$

Q6. √ax + b $$

A.6.

$\begin{array}{l}Put,ax+b=t\\ adx=dt\\ dx=\frac{1}{a}dt\\ I={\displaystyle \int (}ax+b){}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx\\ =\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·\frac{1}{a}dt=\frac{1}{a}\frac{t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}+C\\ =\frac{1}{a}\frac{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\\ =\frac{2}{3a}{(ax+b)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; c\end{array}$

Q7.  $\mathrm{x\surd x\; +\; 2}$

A.7.

$\begin{array}{l}Put,x+2=t\\ dx=dt\\ \Rightarrow x=t-2\\ I={\displaystyle \int \mathrm{x\surd x+2}}dx\\ ={\displaystyle \int (}t-2)\text{\hspace{0.17em}√t}dt\\ =\int \left(t^{1+\frac{3}{2}}-2t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)dt\\ =\int \left(t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)dt\\ =\int t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·dt-{\displaystyle \int 2}·t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·dt\\ =\frac{t^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2\frac{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{2}{5}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\\ =\frac{2}{5}{(x+2)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}{(x+2)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\end{array}$

A.8.

$\begin{array}{l}Put,1+2x^2=t\\ dx2.2x=dt\\ dx.4x=dt\end{array}$

A.10.

$\begin{array}{l}\frac{1}{x-\surd x}=\frac{1}{\surd x(\surd x-1)}\\ Put,\surd x-1=t\\ \frac{1}{\mathrm{2\surd x}}dx=dt\\ I=\int \frac{1}{\surd x(\surd x-1)}dx=\int \frac{2}{t}dt\\ =2log\left|t\right|+c\\ =2log(\surd x-1)+c\end{array}$

Q12.  ${\left(x^3-1\right)}^{\frac{1}{3}}{x}^5$

A.12.

$\begin{array}{l}Put,x^3-1=t\\ 3x^{2}dx=dt\\ =I\int {\left(x^3-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{x}^{5}dx\\ =\int {\left(x^3-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}·x^3·x^2·dx\\ =\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}(t+1)\frac{dt}{3}=\frac{1}{3}\int \left(t^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)dt\\ =\frac{1}{3}\left[\frac{t^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{t^{4/3}}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]+C\\ =\frac{1}{3}\left[\frac{3}{7}{t}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{3}{4}{t}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]+C\\ =\frac{1}{7}{\left(x^3-1\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{1}{4}{\left(x^3-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+C\end{array}$

Q13.  $\frac{x^3}{{\left(2+3x^3\right)}^3}$

A.13.

$\begin{array}{l}Put,2+3x^3=t\\ 9x^{2}dx=dt\\ I=\int \frac{x^3}{{\left(2+3x^3\right)}^3}dx=\frac{1}{9}\int \frac{dt}{t^3}\\ =\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+C\\ =\frac{-1}{18}\left(\frac{1}{t^2}\right)+C=\frac{-1}{18{\left(2+3x^3\right)}^2}+C\end{array}$

Q14.  $\frac{1}{x{(logx)}^m},x>0,m\ne 1$

A.14.

$\begin{array}{l}Put,logx=t\Rightarrow \frac{1}{x}dx=dt\\ I=\int \frac{1}{x{(logx)}^m}dx=\int \frac{dt}{{(t)}^m}\\ =\left(\frac{t^{-m+1}}{1-m}\right)+C=\frac{{(logx)}^{-m+1}}{(1-m)}+C\end{array}$

Q15.  $\frac{x}{9-4x^2}$

A.15.

$\begin{array}{l}Put9-4x^2=t\\ -8xdx=dt\\ xdx=\frac{-1}{8}dt\\ -\frac{1}{8}{\displaystyle \int \frac{dt}{t}}=-\frac{1}{8}log\left|t\right|+c=-\frac{1}{8}log\left|9-4x^2\right|+c\end{array}$

Q16.  $e^{2x+3}$

A.16.

$\begin{array}{l}Put2x+3=t\\ 2dx=dt\\ I=\int e^{2x+3}dx\\ =\frac{1}{2}\int e^t·dt\\ =\frac{1}{2}\left(e^t\right)+C\\ =\frac{1}{2}{e}^{(2x+3)}+C\end{array}$

Q17.  $\frac{x}{e^{x^2}}$

A.17.

$\begin{array}{l}Put{x}^2=t\\ 2xdx=dt\\ I=\int \frac{x}{e^{x^2}}dx=\frac{1}{2}\int \frac{1}{e^t}dt\\ =\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C=\frac{-1}{2}{e}^{-x^2}+C=\frac{-1}{2e^{x^2}}+C\end{array}$

Q18.  $\frac{e^{ta{n}^{-1}x}}{1+x^2}$

A.18.

$\begin{array}{l}Putta{n}^{-1}x=t\\ \frac{1dx}{1+x^2}=dt\\ I=\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx=\int e^{t}dt=e^t+c\\ =e^{ta{n}^{-1}x}+c\end{array}$

Q19.  $\frac{e^{2x}-1}{e^{2x}+1}$

A.19. Dividing both numerator and denominator by e^x, we get

$\begin{array}{l}\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ Put{e}^x+e^{-x}=t\\ \Rightarrow \left(e^x-e^{-x}\right)dx=dt\\ I=\int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx\\ I=\int \frac{dt}{t}=log\left|t\right|+c\\ =log\left|e^x+e^{-x}\right|+c\end{array}$

Q20.  $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

A.20.

$\begin{array}{l}Put{e}^{2x}+e^{-2x}=t\\ \begin{array}{l}\Rightarrow (2e^{2x}-2{e^{-}}^{2x})dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{ll}\Rightarrow 2(e^{2x}-{e^{-}}^{2x})dx=dt\hfill & \hfill \end{array}\\ I=\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx=\int \frac{dt}{2t}=\frac{1}{2}\int \frac{1}{t}dt\\ =\frac{1}{2}log\left|t\right|+=\frac{1}{2}log\left|e^{2x}+e^{-2x}\right|+\end{array}$

Q21.  $ta{n}^2\left(2x-3\right)$

A.21.

$\begin{array}{l}Put2x-3=t\\ 2dx=dt\\ I=\int ta{n}^2(2x-3)dx={\displaystyle \int \left[se{c}^2(2x-3)-1\right]dx}\\ =\frac{1}{2}\int \left(se{c}^{2}t\right)dt-{\displaystyle \int 1}dx\\ =\frac{1}{2}tant-x+C=\frac{1}{2}tan(2x-3)-x+C\end{array}$

Q22.  $se{c}^2\left(7-4x\right)$

A.22.

$\begin{array}{l}Put7-4x=t\\ -4dx=dt\\ I=\int se{c}^2(7-4x)dx\\ =\frac{-1}{4}\int se{c}^{2}tdt=\frac{-1}{4}(tant)+C\end{array}$

A.23.

 

A.24.

$\begin{array}{l}\frac{2cosx-3sinx}{2(3cosx+2sinx)}\\ Put3cosx+2sinx=t\\ \Rightarrow \left(-3sinx+2cosx\right)dx=dt\\ =\int \frac{2cosx-3sinx}{6cosx+4sinx}dx\\ =\int \frac{dt}{2t}=\frac{1}{2}{\displaystyle \int \frac{1}{t}}dt\\ =\frac{1}{2}log\left|t\right|+C\\ =\frac{1}{2}log|3cosx+2sinx|+C.\end{array}$

Q25.  $\frac{1}{co{s}^{2}x{(1-tanx)}^2}$

A.25.

$\begin{array}{l}\frac{se{c}^{2}x}{{(1-tanx)}^2}\\ Put\left(1-tanx\right)=t\\ \Rightarrow se{c}^{2}xdx=dt\\ I=\int \frac{se{c}^{2}x}{{(1-tanx)}^2}dx=\int \frac{-dt}{t^2}\\ =-\int t^{-2}dt\\ =\frac{1}{t}+C\\ =\frac{1}{1-tanx}+C\end{array}$

Q29.  $cotxlogsinx$

A.29.

$\begin{array}{l}Putlogsinx=t\\ \Rightarrow \frac{1}{sinx}·cosxdx=dt\\ \Rightarrow cotxdx=tx\\ I={\displaystyle \int cot}xlogsinxdx\\ ={\displaystyle \int tdt=\frac{t^2}{2}+C=\frac{{\left(logsinx\right)}^2}{2}+C}\end{array}$

Q30.  $\frac{sinx}{1+cosx}$

A.30.

$\begin{array}{l}Put1+cosx=t\\ \Rightarrow -sinxdx=dt\\ I=\int \frac{sinx}{1+cosx}dx={\displaystyle \int -}\frac{dt}{t}\\ =-log\left|t\right|+c\\ =-log\left|1+cosx\right|+c\end{array}$

Q31.  $\frac{sinx}{{(1+cosx)}^2}$

A.31.

$\begin{array}{l}Put1+cosx=t\\ -sinxdx=dt\\ I=\int \frac{sinx}{{(1+cosx)}^2}dx\\ =-\int \frac{dt}{t^2}=-\int t^{-2}dt\\ =\frac{1}{t}+C\\ =\frac{1}{1+cosx}+C\end{array}$

Q32.  $\frac{1}{1+cotx}$

A.32.

$\begin{array}{l}\int \frac{1}{1+\frac{cosx}{sinx}}·dx=\int \frac{1}{\frac{sinx+cosx}{sinx}}dx\\ =\int \frac{sinx}{sin+cosx}·dx=\frac{1}{2}\int \frac{2sinx}{sinx+cosx}dx\\ =\frac{1}{2}\int (\frac{sinx+cosx)+(sinx-cosx)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}(x)+\frac{1}{2}\int \frac{sinx-cosx}{sinx+cosx}dx\\ Putsinx+cosx=t\\ \Rightarrow \left(cosx-sinx\right)dx=dt\\ =\frac{x}{2}+\frac{1}{2}\int \frac{-dt}{t}\\ =\frac{x}{2}-\frac{1}{2}log\left|t\right|+\; c\\ =\frac{x}{2}-\frac{1}{2}log|sinx+cosx|+\; c\end{array}$

Q33.  $\frac{1}{1-tanx}$