Explanation
∫xcos−1(1−x21+x2)dx(x>0)
x=tanθdx=sec2θdθ
I=∫tanθcos−1(cos2θ)sec2θdθ
I=∫2θtanθsec2θdθ
Using by parts
∫u.v dx=u∫v dx −∫(dudx∫v dx)dx
I=2θ∫tanθsec2θdθ−∫2⋅∫tanθsec2θdθ
=2θtan2θ2−2×12∫tan2θdθ
=θtan2θ−[∫(−1+sec2θ)dθ] ................. sec2x=1+tan2x
=θtan2θ+θ−tanθ+c
=θ(1+tan2θ)−tanθ+c
Replacing θ by x,
I=(tan−1x)(1+x2)−x+c
=−x+(1+x2)tan−1x+c
=13[tΨ(t)−∫Ψ(t)dt+C].
∫tan−1(x31+x2)+tan−1(1+x2x3)dx
=∫tan−1[x31+x2+1+x2x31−(x31+x2)(1+x2x3)]dx
=∫tan−1(x6+(1+x2)2(1+x2)(x3)0)dx
=∫π2dx
=π2x+c
∫(elogx+sinx)cosx dx is equal to
We have,
I=∫dx√9−25x2
I=∫dx5√925−x2
I=15∫dx√(35)2−x2
We know that
∫dx√a2−x2=sin−1(xa)+C
Therefore,
I=15sin−1(x35)+C
I=15sin−1(5x3)+C
Hence, this is the correct answer.
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