Show that ∫₀π/3(sec xtan x)/(1+sec²x)dx=tan⁻¹(2)-(π)/(4).

Question 80 of 96

\displaystyle\int_{0}^{\pi/3}\dfrac{\sec x\tan x}{1+\sec^2x}\,dx=\tan^{-1}(2)-\dfrac{\pi}{4}

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Substitutes u = sec x so the numerator becomes du, reducing the integral to the standard arctangent form, then evaluates at the transformed limits.

We need to show $\displaystyle\int_{0}^{\pi/3}\dfrac{\sec x\tan x}{1+\sec^2x}\,dx=\tan^{-1}(2)-\dfrac{\pi}{4}$ .
Let $u=\sec x$ . Then $du=\sec x\tan x\,dx$ , which is exactly the numerator, so the integrand becomes $\dfrac{du}{1+u^2}$ .
Change the limits: when $x=0$ , $u=\sec 0=1$ ; when $x=\pi/3$ , $u=\sec(\pi/3)=2$ .
The integral becomes $\displaystyle\int_{1}^{2}\dfrac{du}{1+u^2}$ . …

Q.Show that $\displaystyle\int_{0}^{\pi/3}\dfrac{\sec x\tan x}{1+\sec^2x}\,dx=\tan^{-1}(2)-\dfrac{\pi}{4}$ .