Ncert Solutions Maths class 12th

Let I = $\int co{s}^{3}x\cdot e^{logsinx}dx.$

$={\displaystyle \int co{s}^{3}xsinxdx}\left\{?e^{logx}=x\right\}$

Putting cos x = t =>-sin x dx = dt =>sin x dx = -dt.

$I\; ={\displaystyle \int t^3(-dt)}=-{\displaystyle \int t^{3}dt}=-\frac{t^4}{4}+C=-\frac{co{s}^{4}x}{4}+C.$

Let I = $\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}dx$

Putting x²= y then,

1 - A (y + 4) + 1B (y + 1)

Comparing the co-efficient,

A + B = 0 _____ (1)

4A + B = 1 ______ (2)

Equation (2) - (1),

4A + B - A - B = 1 - 0

3A = 1 =>A = $\frac{1}{3}.$

and B = - A = -$\frac{1}{3}.$

$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3}\frac{1}{(y+1)}-\frac{1}{3}\frac{1}{(y+4)}$

$=\frac{1}{3}\frac{1}{\left(x^2+1\right)}-\frac{1}{3}\frac{1}{\left(x^2+4\right)}$

$\therefore I=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{dx}{x^2+2^2}$

$=\frac{1}{3}ta{n}^{-1}x-\frac{1}{6}ta{n}^{-1}\frac{x}{2}+\; C$

Let I$=\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}dx$

Putting e^x = t=>e^xdx = dt.

$=\int \frac{dt}{(1+t)(2+t)}=\int \frac{(t+2)-(t+1)}{(t+1)(t+2)}dt$

$=\int \frac{dt}{t+1}-\int \frac{dt}{t+2}$

$=log|t+1|-log|t+2|+C$

$=log\left|\frac{t+1}{t+2}\right|+C.$

$=log\left[\frac{e^x+1}{e^x+2}\right]+C.$

Kindly go through the solution

$I\; ={\displaystyle \int \frac{1}{cos(x+a)cos(x+b)}dx}$

$=\frac{1}{sin(a-b)}\int \frac{sin(a-b)}{cos(x+a)cos(x+b)}·dx$

$=\frac{1}{sin(a-b)}\int \frac{sin[(x+a)-(x+b)]}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}·\int \frac{sin(x+a)cos(x+b)-cos(x+a)sin(x+b)}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}\left[{\displaystyle \int tan(x+a)dx}-{\displaystyle \int tan}(x+b)dx\right]$

$=\frac{1}{sin(a-b)}[-log|cos(x+a)|+log|cos(x+b)|]+\; C$

$=\frac{1}{sin(a-b)}log\left|\frac{cos(x+b)}{cos(x+a)}\right|+\; C$

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{4}x-co{s}^{4}x\right)}{\left(si{n}^{2}x+co{s}^{2}x\right)-si{n}^{2}xco{s}^{2}x-si{n}^{2}xco{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{si{n}^{2}x\left(1-co{s}^{2}x\right)+co{s}^{2}x\left(1-si{n}^{2}x\right)}dx\{\begin{array}{cc}\hfill ?a^2-b^2=(a+b)(a-b)\hfill & \hfill 1=si{n}^{2}x+co{s}^{2}x.\hfill \end{array}$

$={{\displaystyle \int }}^{\text{?}}\frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{4}x\right)(si{n}^{2}x-co{s}^{2}x)\times 1}{si{n}^{2}x\cdot si{n}^{2}x+co{s}^{2}x\cdot co{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{\left(sin4x+co{s}^{4}x\right)}dx$

$={\displaystyle \int -}\left(co{s}^{2}x-si{n}^{2}x\right)dx$

$=-{\displaystyle \int cos}2xdx$

$=-\frac{sin2x}{2}+\; C$

Kindly go through the solution

$\int \frac{e^{5logx}-e^{4logx}}{e^{3logx}-e^{2logx}}dx=\int \frac{e^{log{x}^5}-e^{log{x}^4}}{e^{log{x}^3}-e^{log{x}^2}}dx\{\begin{array}{cc}\hfill ?nlogm\hfill & \hfill =log{m}^n\hfill \end{array}$

$=\int \frac{x^5-x^4}{x^3-x^2}dx\{?e^{logx}=x$

$=\int \frac{x^4(x-1)}{x^2(x-1)}dx=\int x^{2}dx=\frac{x^3}{3}+\; C$

$=\int \frac{sinx}{sin(x-a)}dx$

Let x- a = t=> dx = dt.

$I\; =\int \frac{sin(t+a)}{sint}dt=\int \frac{sintcosa+costsina}{sint}dt$

$=cosa{\displaystyle \int d}t+sina{\displaystyle \int cot}tdt$

$=tcosa+sina·log\left|sint\right|+\; C$

$=(x-a)cosa+sinalog|sin(x-a)|+\; C$

$=xcosa+sinalog|sin(x-a)|+{}_{\mathrm{C1}}$

Where C₁ = C - a cos a

From (1), B = - A = $\left(-\frac{1}{2}\right)=\frac{1}{2}$

From (2) C = 5 - B = 5 $-\frac{1}{2}=\frac{10-1}{2}=\frac{9}{2}$

$\int \frac{5x}{(x+1)\left(x^2+9\right)}dx=\int \left\{\frac{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x+1}+\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.x+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x^2+9}\right\}dx$

$=-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{x}{x^2+9}dx+\frac{9}{2}\int \frac{dx}{x^2+9}$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}\int \frac{9x}{x^2+9}dx+\frac{1}{2}-\int \frac{dx}{x^2+3^2}.$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{9}{2}\times \frac{1}{3}ta{n}^{-1}\frac{x}{3}+\; C$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{3}{2}ta{n}^{-1}\frac{x}{3}+\; C$