# INTEGRATION PROBLEMS.HELP NEEDED

1) Prove that âˆ«x32ax-x2dx=7Ï€a5/8.
2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2).
Upper limit=1/√3;Lower limit=-1/√3.
3) Integrate::: 2-x2/[(1+x)√1-x2].Upper limit=1;Lower limit=0.
4) Integrate:::: log(1+tanx) from 0 to Ï€/2.
5) Integrate::: (√3cos2x-1)/cosx from 0 to Ï€/6.
6) integrate:::: (cos9x)/(cos3x + sin3x) frm 0 to Ï€/2.
7) If f(y)=ey; g(y)=y, y>0 and F(t)=âˆ«f(t-y)g(y)dy from 0 to 1 then F(t)=?
8) Integrate:: [x]f'(x) from 0 to a ,where [.] is the G.I.F. function.
9) Integrate:::::: x4(1-x)4/(1+x2) from 0 to 1.

43
Sayan bisal ·

Manish sir how to do the fourth one

1708
man111 singh ·

\hspace{-16}$For$\bf{(4)::}$Given$\bf{\int_{0}^{\frac{\pi}{2}}\ln(1+\tan \phi)d\phi}$\\\\\\ Let$\bf{\tan \phi = x\;,}$Then$\bf{\sec^2 \phi d\phi = dx\Rightarrow d\phi = \frac{1}{1+x^2}dx}$\\\\\\ And Changing Limit............, We Get\\\\\\ So Integral is$\bf{\int_{0}^{\infty}\frac{\ln(1+x)}{1+x^2}dx = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx+\int_{1}^{\infty}\frac{\ln(1+x)}{1+x^2}dx}$\\\\\\ Now Let$\bf{I=\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx}$and$\bf{J=\int_{1}^{\infty}\frac{\ln(1+x)}{1+x^2}dx}$\\\\\\ Now For Calculation of$\bf{I=\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx}$\\\\\\ Put$\bf{x=\frac{1-t}{1+t}}$and$\bf{dx=-\frac{2}{(1+t)^2}dt}$and Changing Limit\\\\\\ So$\bf{I = \int_{0}^{1}\frac{\ln\left(\frac{2}{1+t}\right)}{1+t^2}dt = \ln(2)\cdot \int_{0}^{1}\frac{1}{1+t^2}dt-\int_{0}^{1}\frac{\ln(1+t)}{1+t^2}dt}$\\\\\\ So$\bf{I = \ln(2)\cdot \frac{\pi}{4}-I\Rightarrow I=\frac{\pi}{8}\cdot \ln(2)}$\\\\\\ \hspace{-18}$Now For Calculation of $\bf{J=\int_{1}^{\infty}\frac{\ln(1+x)}{1+x^2}dx}$\\\\\\ Put $\bf{x=\frac{1}{y}}$ and $\bf{dx = -\frac{1}{y^2}dy}$ and Changing Limit.....\\\\\\ So $\bf{J=\int_{0}^{1}\frac{\ln\left(\frac{1+y}{y}\right)}{1+y^2}dy = \int_{0}^{1}\frac{\ln(1+y)}{1+y^2}dx-\int_{0}^{1}\frac{\ln(y)}{1+y^2}dy}$\\\\\\ So $\bf{J=I+\mathbb{G}\;\;,}$ Where $\bf{\mathbb{G}=-\int_{0}^{1}\frac{\ln(y)}{1+y^2}dy=Catalan\; Constant.}$\\\\\\ So $\bf{\int_{0}^{\frac{\pi}{2}}\ln\left(1+\tan \phi\right)d\phi = I+J=I+I+\mathbb{G}}$\\\\\\ So $\bf{\int_{0}^{\frac{\pi}{2}}\ln\left(1+\tan \phi\right)d\phi = 2\cdot \frac{\pi}{8}\cdot \ln(2)+\mathbb{G}=\frac{\pi}{4}\ln(2)+\mathbb{G}}$\\\\\\

• Sushovan Halder thanks but i didn't understand that G part and catalan constant.From where did u get this name,is it in jee syllabus or beyond that?
• Sushovan Halder plz do the rest ones
187
Swastik Haldar ·

is the answer of 7 is
?

263
Sushovan Halder ·

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1357
Manish Shankar ·

9)

Just divide the numerator by (1+x2) and proceed