Maths Integrals

$\begin{array}{l}2I={\displaystyle {\int }_0^n\frac{\pi tanx}{secx+tanx}dx\Rightarrow 2I=\pi {\displaystyle {\int }_0^n\frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+\frac{sinx}{cosx}}}}dx\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }\frac{sinx+1-1}{1+sinx}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{1}{1+sinx}}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{\left(1-sinx\right)}{\left(1+sinx\right)\left(1-sinx\right)}}dx}\end{array}$

$\begin{array}{l}\Rightarrow 2I=\pi {\left[x\right]}_0^{\pi }-\pi {\displaystyle {\int }_0^{\pi }\frac{1-sinx}{co{s}^{2}x}dx}\\ \Rightarrow 2I={\pi }^2-\pi {\displaystyle {\int }_0^{\pi }\left(se{c}^{2}x-tanxsecx\right)}dx\\ \Rightarrow 2I={\pi }^2-\pi {\left[tanx-secx\right]}_0^{\pi }\end{array}$

$\begin{array}{l}\Rightarrow 2I={\pi }^2-\pi \left[tan\pi -sec\pi -tan0+sec0\right]\\ \Rightarrow 2I={\pi }^2-\pi \left[0-\left(-1\right)-0+1\right]\\ \Rightarrow 2I={\pi }^2-2\pi \\ \Rightarrow 2I=\pi \left(\pi -2\right)\\ I=\frac{\pi }{2}\left(\pi -2\right)\end{array}$

Let I = ${\displaystyle {\int }_0^{\pi /2}sin}2xta{n}^{-1}(sinx)dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}2}sinxcosxta{n}^{-1}(sinx)dx$

Putting sin x = t =>cos xdx = dt.

whenx = 0, t = sin 0 = 0.

$x=\pi /2t=sin\pi /2=1$

? I = ${\displaystyle {\int }_0^{1}2}·tta{n}^{-1}(t)dt$

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

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

= $2\left\{\left[ta{n}^{-1}(1)*\frac{1}{2}-ta{n}^{-1}(0)*\frac{0}{2}\right]-\frac{1}{2}{\displaystyle {\int }_0^1\frac{\left(1+t^2\right)-1}{1+t^2}}dt\right\}$

= $2\left[\frac{\pi }{8}-0\right]-\frac{2}{2}\left\{{\displaystyle {\int }_0^1\frac{1+t^2}{1+t^2}}dt-{\displaystyle {\int }_0^1\frac{dt}{1+t^2}}\right\}$

= $\frac{\pi }{4}-{\displaystyle {\int }_0^{1}d}t+{\left[ta{n}^{-1}t\right]}_0^1$

= $-\frac{\pi }{4}-{[t]}_0^1+\left[ta{n}^{-1}(1)-ta{n}^{-1}(0)\right]$

= $\frac{\pi }{4}-1+\frac{\pi }{4}=2*\frac{\pi }{4}-1=\frac{\pi }{2}-1$

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinx+cosx}{9+16si{n}^{2}x}}dx$

Let sin x – cos x = t. =>(cosx + sin x) dx = dt.

and (sin x – cos x)² = t²

sin²x + cos²x – 2 sin x cos x = t²

1 – sin2x = t².

sin2t = 1 - t².

When x = 0, t = sin 0 – cos 0 = –1$$

? I = ${\displaystyle {\int }_{-1}^0\frac{dt}{9+16\left(1-t^2\right)}}={\displaystyle {\int }_{-1}^0\frac{dt}{9+16-16t^2}}$

= ${\displaystyle {\int }_{-1}^0\frac{dt}{25-16t^2}}$

= ${\displaystyle {\int }_{-1}^0}\frac{dt}{16\left(\frac{25}{16}-t^2\right)}$

= $\frac{1}{16}{\displaystyle {\int }_{-1}^0\frac{dt}{{\left(\frac{5}{4}\right)}^2-t^2}}$

= $\frac{1}{16}{\left[\frac{1}{2*\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}log\left|\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+t}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-t}\right|\right]}_{-1}^0\left\{{\displaystyle \int \frac{dx}{a^2-x^2}=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|}\right\}$

= $\frac{1}{16}*\frac{4}{2*5}{\left[log\left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0$

= $\frac{1}{40}\left[log\frac{5+4*0}{5-4*0}-log\frac{5+4(-1)}{5-4(-1)}\right]$

= $\frac{1}{40}\left[log\frac{5}{5}-log\frac{1}{9}\right]$

= $\frac{-1}{40}\left[log1-log{9}^{(-1)}\right]$

= $\frac{1}{40}[0-(-1)log9]$

= $\frac{1}{40}log9$

= $\frac{1}{40}log{3}^2=\frac{2}{40}log3$

= $\frac{1}{20}log3$

Kindly go through the solution

Kindly go through the solution

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\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*\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 = $\frac{-\pi }{6}+\frac{2}{3}*\frac{\pi }{2}=-\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{6}.$

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]$

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

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

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

Kindly go through the solution

Kindly go through the solution