5)∫cot−1ex⋅e−xdx
−∫tan−1e−xde−x
Applying integration by parts,
−e−xtan−1(e−x)+21ln∣e−2x+1∣+C
21ln∣e2x+1∣−x−cot−1(ex)e−x+C
Ans-C
5)∫cot−1ex⋅e−xdx
−∫tan−1e−xde−x
Applying integration by parts,
−e−xtan−1(e−x)+21ln∣e−2x+1∣+C
21ln∣e2x+1∣−x−cot−1(ex)e−x+C
Ans-C
∫0√lnπ/2cos(ex2)2xex2dx=∫1π/2cos(ex2)dex2
1−sin1
4) f(g(x))=x ...(1)
f'(g(x))g'(x)=1
g'(0)=1f'(g(0))
From(1)
f(g(0))=0
f(x) is zero when both upper & lower limit are equal.
f'(x)= 1√(1+x4)
f'(2)=1√17
Hence g'(0)=√17
(c)