bitsat 2009

1) take x=tanθ

why r u not using biparts taking u= 1 and v= tan-1 x

1)
use by parts
\int_{0}^{1}{tan^{-1}x}=x.tan^{-1}x|_{0}^{1}-\int_{0}^{1}{\frac{x}{1+x^{2}}}

i think u r facing trouble to solve it
multiply and divide it by 2
then take
1+x²=t
2xdx=dt

so u get
ln(1+x²)|₀¹

=pi/4-ln2

Q 3 and Q 4 are same .Just take a common in 4)
for Q3)
u see that
∫1-cosx/sin²x dx
∫cosec²x-∫cotxcosecxdx
=-cotx+cosecx

and 4 the 4) u will get 1/a(-cotx+cosecx)