Evaluate: $\int_0^1 \dfrac{1}{1+x^2}\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx$

Evaluate: ∫₀¹ (1)/(1+x²)sin⁻¹((2x)/(1+x²))dx

Miscellaneous Exercise 4 · Q71

\int_0^1 \dfrac{1}{1+x^2}\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx

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Let $x=\tan\theta,\ dx=\sec^2\theta\,d\theta$ , $1+x^2=\sec^2\theta$ . Then $\dfrac{2x}{1+x^2}=\dfrac{2\tan\theta}{\sec^2\theta}=2\sin\theta\cos\theta=\sin2\theta$ , so $\sin^{-1}\!\big(\tfrac{2x}{1+x^2}\big)=\sin^{-1}(\sin2\theta)=2\theta$ (valid since $2\theta\in[0,\pi/2]$ for $x\in[0,1]$ ). Also $\dfrac{dx}{1+x^2}=d\theta$ . …

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Q.Evaluate: $\int_0^1 \dfrac{1}{1+x^2}\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx$

Concept understanding — Evaluation of Definite Integrals by Substitution

Q.Evaluate: ∫0111+x2sin⁡−1 ⁣(2x1+x2)dx\int_0^1 \dfrac{1}{1+x^2}\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx∫01​1+x21​sin−1(1+x22x​)dx

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Q.Evaluate: $\int_0^1 \dfrac{1}{1+x^2}\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx$

Unlock everything free for 14 days