Integration @ 01

Calculus Level 1

E v a l u a t e ∫ arcsin ⁡ x d x . Evaluate\quad \int { \arcsin { x\quad dx. } }

x arcsin ⁡ x + 1 − x 2 x\arcsin { x } +\sqrt { 1-{ x }^{ 2 } } x sin ⁡ x + 1 − x 2 { x }\sin { x } +\sqrt { 1-{ x }^{ 2 } } x 2 sin ⁡ x + 1 − x 2 { x }^{ 2 }\sin { x } +\sqrt { 1-{ x }^{ 2 } } arcsin ⁡ x + 1 − x 2 \arcsin { x } +\sqrt { 1-{ x }^{ 2 } } arcsin ⁡ x + x 1 − x 2 \arcsin { x } +x\sqrt { 1-{ x }^{ 2 } }

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Saurav Pal by Saurav Pal , 26, India

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1 solution

Naren Bhandari
Oct 5, 2017

I = ∫ sin ⁡ − 1 x d x I = \int \sin ^{-1}x\,dx let us substitute sin ⁡ − 1 x = u ⇒ x = sin ⁡ u \sin^{-1}x = u \Rightarrow x = \sin u Differentiating we get 1 1 − x 2 d x = d u ⇒ d x = 1 − x 2 d u = cos ⁡ u d u \frac{1}{\sqrt{1-x^2}}\,dx = \,du \Rightarrow \,dx = \sqrt{1-x^2}\,du =\cos u\,du I = ∫ sin ⁡ − 1 d x ⇒ ∫ u c o s u d u I = \int \sin^{-1}\,dx \Rightarrow \int u \ cos u\,du

I = u ∫ sin ⁡ u − ∫ ( d ( u ) d u ∫ sin ⁡ u d u ) d u I = u\int \sin u - \int\left(\frac{\,d(u)}{\,du}\int \sin u\,du\right)\,du = u sin ⁡ u − ∫ sin ⁡ u d u = u \sin u - \int \sin u\,du = x sin ⁡ − 1 + cos ⁡ u + C = x \sin^{-1} + \cos u + C = x sin ⁡ − 1 + 1 − x 2 + C = x \sin^{-1} +\sqrt{1-x^2} +C

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