Ncert Solutions Maths class 12th

Let I = ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left(\frac{1-sinx}{1-cosx}\right)dx.$

= ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[\frac{1-2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2si{n}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx.$

= ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[\frac{1}{2}cose{c}^2\frac{x}{2}-cot\frac{x}{2}\right]dx$

= – ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[cot\frac{x}{2}-\frac{1}{2}cose{c}^2\frac{x}{2}\right]dx$

= $-{\left[e^x·cot\frac{x}{2}\right]}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }$ $\{?\int e^x\left[f(x)+f^{\prime }(x)\right]dx=e^{n}f(x)$

= $-\left[e^{\pi }cot\frac{x}{2}-\frac{\pi }{e^2}cot\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\right]$

= $-\left[e^{\pi }\times 0-e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·cot\frac{1}{4}\right]$

= $0+e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 1.$

= $e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Kindly go through the solution

Kindly go through the solution

Let $I={{\displaystyle \int }}^{\text{?}}\frac{x^2+x+1}{{(x+1)}^2(x+2)}dx$

The integrate is of form,

$\frac{x^2+x+1}{{(x+1)}^2(x+2)}=\frac{A}{x+1}+\frac{B}{{(x+1)}^2}+\frac{C}{x+2}.$

$\Rightarrow x^2+x+1=A(x+1)(x+2)+B(x+2)+C{\left(x+1\right)}^2.$

$=A\left(x^2+3x+2\right)+B(x+2)+C\left(x^2+1+2x\right)$

Comparing the co-efficients,

A + C = 1 .(1)

3A + B + C = 1 .(2)

2A + 2B + C = 1 .(3)

Equation. (2) – 2 × Equation (1),

$3A+B+2C-2A-2C=1-2\times 1.$

$\Rightarrow$ A + B = –1 .(4)

Equation (3) - (1),

2A + 2B + C = A – C = 1 – 1

$\Rightarrow$ A + 2B = 0 .(5)

Equation (5) - Equation (4),

A + 2B - A - B = 0 - (-1)

$\Rightarrow$ B = 1.

From (4),

Let I = ${{\displaystyle \int }}^{\text{?}}\frac{2+sin2x}{1+cos2x}{e}^{x}dx\cdot$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[\frac{2+2sinxcosx}{2co{s}^{2}2}\right]dx\left\{\begin{array}{ll}\therefore sin2\theta =2sin\theta cos\theta \hfill & cos2\theta =2co{s}^2\theta -1\hfill \end{array}\right\}$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[\frac{1}{co{s}^{2}x}+\frac{sinx}{cosx}\right]dx$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[tanx+se{c}^{2}x\right]dx$ is in the form

${{\displaystyle \int }}^{\text{?}}{e}^x\left[f(x)+f^{\prime }(x)\right]dx$

$f(x)=tanx$

$f^{\prime }(x)=se{c}^{2}x.$

$\therefore I=e^{x}f(x)+c=e^{x}tanx+c.$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Let I $={\displaystyle ?f^{?}(ax+b)·[f]{(ax+b)}^n}dx.$

$=\frac{1}{a}{\displaystyle ?[f]{(ax+b)}^n*\left[a{f}^{?}(ax+b)\right]}dx.$

Putting f (ax + b) = t.

$f^{?}(ax+b)\frac{d}{dx}(ax+b)=\frac{dt}{dx}$

af'  (ax + b) dx = dt

$?=\frac{1}{a}{\displaystyle ?t^{x}dt}$

$=\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]$

$=\frac{1}{a(x+1)}[f]{(ax+b)}^{n+1}+\; C$

Let I$={\displaystyle \int e^{3logx}{\left(x^4+1\right)}^{-1}dx}$

$={\displaystyle \int \frac{e^{log{x}^3}}{\left(x^4+1\right)}dx}$

$={\displaystyle \int \frac{x^3}{x^4+1}}dx=\frac{1}{4}{\displaystyle \int \frac{4x^3}{x^4+1}}dx$

$=\frac{1}{4}log\left|x^4+1\right|+\; C$