1why r u not using biparts taking u= 1 and v= tan-1 x
8 Answers
111)
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
112)
multiply and divide the eq by 2
u get
1/2[∫cos5x + ∫cosx]
hence
1/2[sin5x/5+sinx]
11Q 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)