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Mathematics Question on Methods of Integration

The integral ∫0π213+2sin⁡x+cos⁡x dx\int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx is equal to

A

tan⁡−1(2)\tan^{-1}(2)

B

tan⁡−1(2)−π4\tan^{-1}(2) - \frac{\pi}{4}

C

12tan⁡−1(2)−π8\frac{1}{2}\tan^{-1}(2) - \frac{\pi}{8}

D

12\frac{1}{2}

Answer

tan⁡−1(2)−π4\tan^{-1}(2) - \frac{\pi}{4}

Explanation

Solution

I=∫0π213+2sin⁡x+cos⁡x dxI = \int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx

=∫0π21+tan⁡2(x2)3(1+tan⁡2(x2))+2(2tan⁡(x2))+(1−tan⁡2(x2)) dx=\int_0^{\frac{\pi}{2}} \frac{1 + \tan^2\left(\frac{x}{2}\right)}{3\left(1 + \tan^2\left(\frac{x}{2}\right)\right) + 2\left(2\tan\left(\frac{x}{2}\right)\right) + \left(1 - \tan^2\left(\frac{x}{2}\right)\right)} \, dx

Let tan⁡(x2)=t⇒sec⁡2(x2) dx=2 dt\tan\left(\frac{x}{2}\right) = t \quad \Rightarrow \quad \sec^2\left(\frac{x}{2}\right) \, dx = 2 \, dt

I=∫012dt4+2t2+4tI = \int_0^1 \frac{2dt}{4 + 2t^2 + 4t}

I=∫01dtt2+2t+2I = \int_0^1 \frac{dt}{t^2 + 2t + 2}

I=∫01dt(t+1)2+1I = \int_0^1 \frac{dt}{(t+1)^2 + 1}

I=tan⁡−1(t+1)∣01I = \tan^{-1}(t+1) \Big|_{0}^{1}

=I=tan⁡−1(2)−π4=I = \tan^{-1}(2) - \frac{\pi}{4}
So, the correct option is (B): tan⁡−1(2)−π4\tan^{-1}(2) - \frac{\pi}{4}