Solveeit Logo
Login

Question

Question: Evaluate : \(\int_{- \pi/3}^{–\pi/6}\frac{dx}{1 + \tan^{2n}x}\)...

Evaluate : ∫−π/3–π/6dx1+tan⁡2nx\int_{- \pi/3}^{–\pi/6}\frac{dx}{1 + \tan^{2n}x}

A

π6\frac{\pi}{6}

B

π12\frac{\pi}{12}

C

π3\frac{\pi}{3}

D

None of these

Answer

π12\frac{\pi}{12}

Explanation

Solution

Let I = ∫−π/3−π/6dx1+tan⁡2nx\int_{- \pi/3}^{- \pi/6}\frac{dx}{1 + \tan^{2n}x}

= ∫π/6π/3dx1+(tan⁡(–x))2n\int_{\pi/6}^{\pi/3}\frac{dx}{1 + (\tan(–x))^{2n}} (By Prop. X)

= ∫π/6π/3cos⁡2nxdx(sin⁡2nx+cos⁡2nx)\int_{\pi/6}^{\pi/3}\frac{\cos^{2n}xdx}{(\sin^{2n}x + \cos^{2n}x)} … (1) and

I = ∫π/6π/3cos⁡2n(π3+π6−x)dxsin⁡2n(π3+π6−x)+cos⁡2n(π3+π6−x)\int_{\pi/6}^{\pi/3}\frac{\cos^{2n}\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)dx}{\sin^{2n}\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right) + \cos^{2n}\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}

(By Prop. V)

= ∫π/6π/3sin⁡2nxdxcos⁡2nx+sin⁡2nx\int_{\pi/6}^{\pi/3}\frac{\sin^{2n}xdx}{\cos^{2n}x + \sin^{2n}x} … (2)

Adding (1) & (2) we get

2I = ∫π/6π/31\int_{\pi/6}^{\pi/3}1. dx = π3\frac{\pi}{3}– π6\frac{\pi}{6} = π6\frac{\pi}{6}

\ I = π12\frac{\pi}{12}