Integrals

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4si{n}^{2}x}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4\left(1-co{s}^{2}x\right)}}dx\{?1=co{s}^{2}x+si{n}^{2}x.$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4-4co{s}^{2}x}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{-3co{s}^{2}x}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\left(4-3co{s}^{2}x\right)-4}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}\left[{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4}{4-3co{s}^{2}x}}dx\right]$

= $-\frac{1}{3}\left\{{[x]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4}{4-3\times \frac{1}{se{c}^{2}x}}}dx\right\}$

= $-\frac{1}{3}\left\{\frac{\pi }{2}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4se{c}^{2}x}{4se{c}^{2}x-3}}dx\right\}$

= $-\frac{1}{3}\left\{\frac{\pi }{2}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4se{c}^{2}x}{4\left(1+ta{n}^{2}x\right)-3}}dx\right\}\left\{se{n}^{2}x=1+ta{n}^{2}x\right\}$

= $\frac{-\pi }{6}+\frac{2}{3}{}_1$

where I₁ = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{2ta{n}^{2}x}{1+4ta{n}^{2}x}}dx.$

Putting 2tan x = t =>2 sin²xdx = dt& when x = 0, t = 2 tan (0) = 0

x = $\frac{\pi }{2}$ , t = 2 tan $\frac{\pi }{2}$ = ∞

I1 = ${\displaystyle {\int }_0^{\infty }\frac{dt}{1+t^2}}.$

= $[ta{n}^{-}t]{}_0^{\infty }$

= tan^–1 (∞) – tan^–10

= $\frac{\pi }{2}-0$ = $\frac{\pi }{2}.$

? I = $\frac{-\pi }{6}+\frac{2}{3}\times \frac{\pi }{2}=-\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{6}.$

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinxcosx}{co{s}^{4}x+si{n}^{4}x}}dx.$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinxcosx}{co{s}^{4}x\left(1+\frac{si{n}^{4}x}{co{s}^{4}x}\right)}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{\frac{sinx}{cosx}·\frac{cosx}{cosx}\frac{1}{co{s}^{2}x}}{\left(1+ta{n}^{4}x\right)}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{tanx·se{c}^{2}x}{1+ta{n}^{4}x}}dx$

= $\frac{1}{2}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{2tanx·se{c}^{2}x}{1+ta{n}^{4}x}}dx.$

Putting tan²x = t =>2 tan x?sec²xdx = dt

When x = 0, t = tan²x = tan² 0 = 0

x = $\frac{\pi }{4}$ t = tan² $\frac{\pi }{4}$ = 1² = 1.

? I = $\frac{1}{2}{\displaystyle {\int }_0^1\frac{dt}{1+t^2}}=\frac{1}{2}{\left[ta{n}^{-1}t\right]}_0^1$

= $\frac{1}{2}\left[ta{n}^{-1}(1)-ta{n}^{-1}(0)\right]$

= $\frac{1}{2}\left[\frac{\pi }{4}-0\right]=\frac{\pi }{8}$

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