Solveeit Logo
Login

Question

Question: $\int \frac{\sin 2x}{1+\cos^4 x} dx$ is equal to...

∫sin⁡2x1+cos⁡4xdx\int \frac{\sin 2x}{1+\cos^4 x} dx is equal to

A

cos⁡−1(cos⁡2x)+c\cos^{-1}(\cos^2 x) + c

B

sin⁡−1(cos⁡2x)+c\sin^{-1}(\cos^2 x) + c

C

cot⁡−1(cos⁡2x)+c\cot^{-1}(\cos^2 x) + c

D

none of these

Answer

cot⁡−1(cos⁡2x)+c\cot^{-1}(\cos^2 x) + c

Explanation

Solution

The given integral is ∫sin⁡2x1+cos⁡4xdx\int \frac{\sin 2x}{1+\cos^4 x} dx.

Let's use a substitution method. Let t=cos⁡2xt = \cos^2 x. Then dtdx=−2sin⁡xcos⁡x=−sin⁡2x\frac{dt}{dx} = -2 \sin x \cos x = -\sin 2x. This implies dt=−sin⁡2x dxdt = -\sin 2x \, dx, or sin⁡2x dx=−dt\sin 2x \, dx = -dt.

Now substitute tt and dtdt into the integral: ∫sin⁡2x1+cos⁡4xdx=∫−dt1+t2=−∫11+t2dt=−tan⁡−1t+C\int \frac{\sin 2x}{1+\cos^4 x} dx = \int \frac{-dt}{1+t^2} = -\int \frac{1}{1+t^2} dt = -\tan^{-1} t + C.

Now, substitute back t=cos⁡2xt = \cos^2 x: −tan⁡−1(cos⁡2x)+C-\tan^{-1}(\cos^2 x) + C.

We know the trigonometric identity relating tan⁡−1x\tan^{-1} x and cot⁡−1x\cot^{-1} x: tan⁡−1x+cot⁡−1x=π2\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}. From this, we can write tan⁡−1x=π2−cot⁡−1x\tan^{-1} x = \frac{\pi}{2} - \cot^{-1} x. Applying this to our result: −tan⁡−1(cos⁡2x)+C=−(π2−cot⁡−1(cos⁡2x))+C=−π2+cot⁡−1(cos⁡2x)+C-\tan^{-1}(\cos^2 x) + C = - \left( \frac{\pi}{2} - \cot^{-1}(\cos^2 x) \right) + C = -\frac{\pi}{2} + \cot^{-1}(\cos^2 x) + C. Since CC is an arbitrary constant of integration, and −π2-\frac{\pi}{2} is also a constant, we can combine them into a new arbitrary constant, say C′C'. So, the result is: cot⁡−1(cos⁡2x)+C′\cot^{-1}(\cos^2 x) + C'.