# a few problems by goyalji

Integrate:
1) (1+x-2/3)/(1+x)
2)√cos2x/sinx
3)(x2 + n(n-1))/(xsinx+ncosx)2

213
Nandikesh Singh ·

For third one multiply and divide x^(2n-2).

• Sushovan Halder @nandikesh:then what to do?
1357
Manish Shankar ·

first one take x=t3

1357
Manish Shankar ·

second cos2x=t2 might help

187
Swastik Haldar ·

after multiplying and dividing by x^(2n-2),
I multiplied and divided by cos x. then rearranging, I got,

1708
man111 singh ·

\hspace{-16}\bf{(3)\;::\; }$Given$\bf{\int\frac{x^2+n(n-1)}{\left(x\cdot \sin x+n\cdot \cos x\right)^2}dx}$\\\\\\ Now$\bf{x\cdot \sin x+n\cdot \cos x = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}}$\\\\\\ Now Let$\bf{\frac{n}{\sqrt{x^2+n^2}}=\cos \phi}$and$\bf{\frac{x}{\sqrt{x^2+n^2}}=\sin \phi}$\\\\\\ So$\bf{\tan \phi = \frac{x}{n}\Rightarrow \tan \phi = \tan^{-1}\left(\frac{x}{n}\right)}$\\\\\\ So$\bf{x\cdot \sin x+n\cdot \cos x = \sqrt{x^2+n^2}\cdot \cos \left(x-\phi\right)}$\\\\\\ So Integral is Convert into$\bf{\int\frac{1}{\cos^2\left(x-\phi\right)}\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx}$\\\\\\ So Integral is$\bf{\int \sec^2 \left(x-\phi\right)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx}$\\\\\\ Let$\bf{(x-\phi) = t\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=t\;,}$Then$\bf{\left(1-\frac{n^2}{x^2+n^2}\cdot \frac{1}{n}\right)dx = dt}$\\\\\\ \hspace{-20}$ So $\bf{\left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dt}$\\\\\\ So Integration is convert into $\bf{\int \sec^2\left(t\right)dt = \tan (t)+\mathbb{C}}$\\\\\\ $\bf{\Rightarrow \tan \left(x-\phi\right)+\mathbb{C}=\tan (x-\tan^{-1}\left(\frac{x}{n}\right))+\mathbb{C}}$\\\\\\ $\bf{\Rightarrow \frac{\tan x-\frac{x}{n}}{1+\tan x\cdot \frac{x}{n}}+\mathbb{C}=\left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathbb{C}}$\\\\\\ So $\bf{\int\frac{x^2+n(n-1)}{\left(x\cdot \sin x+n\cdot \cos x\right)^2}=\left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathbb{C}}$

• Sushovan Halder lot of thanks to both of u
43
Sayan bisal ·

For second one

1708
man111 singh ·

\hspace{-16}$For$\bf{(2)}:\; $Let$\bf{I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\sqrt{\cos^2 x-\sin^2 x}}{\sin x}dx}$\\\\\\ So$\bf{I=\int \sqrt{\cot^2 x-1}dx=\int\frac{\cot^2 x-1}{\sqrt{\cot^2 x-1}}dx = \int\frac{\csc^2 x}{\sqrt{\cot^2 x-1}}dx-2\int\frac{1}{\sqrt{\cot^2 x-1}}dx}$\\\\\\ So$\bf{I = -\int\frac{(\cot x)^{'}}{\sqrt{\cot^2 x-1}}dx+2\int\frac{(\cos x)^{'}}{\sqrt{2\cos^2 x-1}}dx.............}\$

43
Sayan bisal ·

For the first one the answer is 2log(1+x)+c