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].
\int {{{\tan }^{ - 1}}\left( {{{{x^3}} \over {1 + {x^2}}}} \right) + } {\tan ^{ - 1}}\left( {{{1 + {x^2}} \over {{x^3}}}} \right)dx
= \int {{{\tan }^{ - 1}}\left[ {{{{{{x^3}} \over {1 + {x^2}}} + {{1 + {x^2}} \over {{x^3}}}} \over {1 - \left( {{{{x^3}} \over {1 + {x^2}}}} \right)\left( {{{1 + {x^2}} \over {{x^3}}}} \right)}}} \right]} dx
= \int {{{\tan }^{ - 1}}\left( {{{{{{x^6} + {{\left( {1 + {x^2}} \right)}^2}} \over {\left( {1 + {x^2}} \right)\left( {{x^3}} \right)}}} \over 0}} \right)} dx
= \int {{\pi \over 2}dx}
= {\pi \over 2}x + c
\int ( e^{\log x} + \sin x) \cos x \ dx is equal to
We have,
I=\int{\dfrac{dx}{\sqrt{9-25{{x}^{2}}}}}
I=\int{\dfrac{dx}{5\sqrt{\dfrac{9}{25}-{{x}^{2}}}}}
I=\dfrac{1}{5}\int{\dfrac{dx}{\sqrt{{{\left( \dfrac{3}{5} \right)}^{2}}-{{x}^{2}}}}}
We know that
\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C
Therefore,
I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{x}{\dfrac{3}{5}} \right)+C
I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C
Hence, this is the correct answer.
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