1replaxe x-> -x
=> 2I = ∫cos2x
you can do it frm here
17 Answers
1708\hspace{-16}\int\sin^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$\\\\\\ $\int \sin^{-1}\left(\frac{2x+2}{\sqrt{(2x+2)^2+3^2}}\right)dx$\\\\\\ Put $2x+2=t\Leftrightarrow 2dx=dt\Leftrightarrow dx=\frac{1}{2}dt$ \\\\\\$\int \sin^{-1}\left(\frac{t}{\sqrt{t^2+3^2}}\right)dt$\\\\\\ Put $t=3\tan \theta\Leftrightarrow dt=3\sec^2 \theta d\theta$\\\\\\ $\int \sin^{-1}\left(\frac{3\tan \theta}{3\sec \theta}\right).3\sec^2 \theta d\theta$\\\\\\ $3.\int \sin^{-1}(\sin \theta).\sec^2 \theta d\theta=3\int \theta.\sec^2 \theta d\theta$\\\\\\ Using \text{I.B.P}\\\\ $3.\left\{\theta.\tan \theta -\ln \mid \sec \theta\mid \right\}+C$\\\\ $3.\left\{\tan^{-1}\left(\frac{t}{3}\right).\left(\frac{t}{3}\right)-\frac{1}{2}\ln \mid 1+\frac{t^2}{9}\mid \right\}+C$\\\\ Now Put $t=2x+2$
1708\hspace{-16}\int\sqrt{\frac{1+x^2}{1+\sqrt{1+x^2}}}dx$\\\\\\ Put $1+x^2=t^2\Leftrightarrow 2xdx=2tdt\Leftrightarrow dx=\frac{t}{\sqrt{t^2-1}}dt$\\\\\\ $\int\frac{t}{\sqrt{1+t}}.\frac{t}{\sqrt{t^2-1}}dt=\int\frac{t^2}{(t+1).\sqrt{t-1}}dt$\\\\\\ Put $t-1=u^2\Leftrightarrow dt=2udu$\\\\\\ $\int\frac{(u^2+1)^2.2u}{(u^2+2).u}du=2.\int\frac{(\underbrace{u^2+2}-1)^2}{(u^2+2)}du$\\\\\\ $=2.\int\frac{(u^2+2)^2+1-2(u^2+2)}{(u^2+2)}du$\\\\\\ $=2.\int(u^2+2).du+2.\int\frac{1}{u^2+(\sqrt{2})^2}du-4.\int 1.du\\\\\\ =2.\frac{u^3}{3}+4u-2.\frac{1}{\sqrt{2}}.\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)-4u+C$\\\\\\ $=\frac{2u^3}{3}-\sqrt{2}.\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C\\\\\\ =\frac{2.(t-1)^{\frac{3}{2}}}{3}-\sqrt{2}.\tan^{-1}\left(\frac{\sqrt{t-1}}{\sqrt{2}}\right)+C$\\\\\\ $=\frac{2.(\sqrt{1+x^2}-1)^{\frac{3}{2}}}{3}-\sqrt{2}.\tan^{-1}\left(\frac{\sqrt{\sqrt{1+x^2}-1}}{\sqrt{2}}\right)+C$$
1i got the solution bt what if i put the lower limit 0 n upper limit infinity?
1you know you can do it easily if the numerator were the denominator nd the denominator were the numerator..so make it!
i mean take x-pi4 = t and expant using sin(A+B) formula
1what would be the value of
∫cos2x/(1+ax)dx , a>0 limit extends from -pi to +pi
1708\hspace{-16}\int\frac{\sec^2 x}{\left(\sec x+\tan x\right)^n}dx$\\\\\\ $=\int\frac{\sec x.\sec x.(\sec x+\tan x)}{(\sec x+\tan x)^{n+1}}dx$\\\\\\ Now Let $(\sec x+\tan x)=t\Leftrightarrow \sec x.(\sec x+\tan x)dx=dt$\\\\\\ and Using $(\sec^2 x-\tan ^2 x)=1\Leftrightarrow (\sec x-\tan x).(\sec x+\tan x)=1$\\\\\\ So $(\sec x-\tan x)=\frac{1}{t}$\\\\\\ so $\sec x=\frac{1}{2}.\left(t+\frac{1}{t}\right)=\frac{(t^2+1)}{2t}$\\\\\\ So $\int\frac{t^2+1}{t^{n+1}.2t}dt=\int\frac{t^2+1}{t^n.2.t^2}dt=\frac{1}{2}.\int t^{-n}dt+\frac{1}{2}.\int t^{-(n+2)}dt$\\\\\\ $=\frac{t^{(1-n)}}{2.(1-n)}-\frac{t^{-(n+1)}}{2.(n+1)}+C$\\\\\\ $=\frac{(\tan x+\sec x)^{(1-n)}}{2.(1-n)}-\frac{(\tan x+\sec x)^{-(n+1)}}{2(n+1)}+C$
1708\hspace{-16}\int\frac{1}{x.(a+bx^n)^2}dx\\\\\\ $Put $(a+bx^n)=\frac{1}{t}\Leftrightarrow n.bx^{n-1}dx=-\frac{1}{t^2}dt$\\\\\\ $n.\left(\frac{1}{t}-a\right).\frac{dx}{x}=-\frac{1}{t^2}dt\Leftrightarrow \frac{dx}{x}=-\frac{1}{t^2}.\frac{1}{n}.\frac{t}{(1-at)}dt=-\frac{1}{n}.\frac{1}{t.(1-at)}$\\\\\\ So Integral is $-\frac{1}{n}\int\frac{t^2}{t.(1-at)}dt=\frac{1}{n}.\int\frac{t}{(at-1)}dt$\\\\\\ $=\frac{1}{a.n}.\int\frac{(at-1)+1}{(at-1)}dt=\frac{1}{a.n}\int 1.dt+\frac{1}{a.n}.\int\frac{1}{(at-1)}dt$\\\\\\ $=\frac{1}{a.n}t+.\frac{1}{a.n}.\ln \left|at-1\right|+C$\\\\\\ $=\frac{1}{a.n}.\frac{1}{(a+bx^n)}+\frac{1}{a.n}.\ln\left|\frac{a}{(a+bx^n)}-1\right|+C$