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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

(c) 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.

Step 1: Simplify the integrand using trigonometric identities. We know that sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x. So, the integral becomes: ∫2sin⁡xcos⁡x1+cos⁡4xdx\int \frac{2 \sin x \cos x}{1+\cos^4 x} dx

Step 2: Use substitution method. Let u=cos⁡2xu = \cos^2 x. To find dudu, differentiate uu with respect to xx: dudx=ddx(cos⁡2x)\frac{du}{dx} = \frac{d}{dx}(\cos^2 x) Using the chain rule, ddx(f(x)n)=nf(x)n−1f′(x)\frac{d}{dx}(f(x)^n) = n f(x)^{n-1} f'(x). Here, f(x)=cos⁡xf(x) = \cos x and n=2n=2. dudx=2(cos⁡x)2−1⋅ddx(cos⁡x)\frac{du}{dx} = 2 (\cos x)^{2-1} \cdot \frac{d}{dx}(\cos x) dudx=2cos⁡x(−sin⁡x)\frac{du}{dx} = 2 \cos x (-\sin x) dudx=−2sin⁡xcos⁡x\frac{du}{dx} = -2 \sin x \cos x So, du=−2sin⁡xcos⁡x dxdu = -2 \sin x \cos x \, dx. This implies 2sin⁡xcos⁡x dx=−du2 \sin x \cos x \, dx = -du.

Step 3: Substitute into the integral. Substitute u=cos⁡2xu = \cos^2 x and 2sin⁡xcos⁡x dx=−du2 \sin x \cos x \, dx = -du into the integral: ∫−du1+u2\int \frac{-du}{1+u^2} =−∫11+u2du= -\int \frac{1}{1+u^2} du

Step 4: Integrate the standard form. The integral of 11+u2\frac{1}{1+u^2} is tan⁡−1(u)\tan^{-1}(u). =−tan⁡−1(u)+C= -\tan^{-1}(u) + C where CC is the constant of integration.

Step 5: Substitute back the original variable. Substitute u=cos⁡2xu = \cos^2 x back into the result: =−tan⁡−1(cos⁡2x)+C= -\tan^{-1}(\cos^2 x) + C

Step 6: Convert to match the options (if necessary). We know the trigonometric identity tan⁡−1(x)+cot⁡−1(x)=π2\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}. From this, we can write −tan⁡−1(x)=cot⁡−1(x)−π2-\tan^{-1}(x) = \cot^{-1}(x) - \frac{\pi}{2}. Applying this identity to our result: −tan⁡−1(cos⁡2x)+C=cot⁡−1(cos⁡2x)−π2+C-\tan^{-1}(\cos^2 x) + C = \cot^{-1}(\cos^2 x) - \frac{\pi}{2} + C Since CC is an arbitrary constant, C−π2C - \frac{\pi}{2} is also an arbitrary constant. Let's denote it as C′C'. =cot⁡−1(cos⁡2x)+C′= \cot^{-1}(\cos^2 x) + C'

Comparing this with the given options, option (c) matches our result.

The final answer is (c)\boxed{\text{(c)}}.

Explanation of the solution: The integral ∫sin⁡2x1+cos⁡4xdx\int \frac{\sin 2x}{1+\cos^4 x} dx is solved by substitution. Let u=cos⁡2xu = \cos^2 x. Then du=−2sin⁡xcos⁡xdx=−sin⁡2xdxdu = -2 \sin x \cos x dx = -\sin 2x dx. The integral transforms to ∫−du1+u2\int \frac{-du}{1+u^2}, which integrates to −tan⁡−1(u)+C-\tan^{-1}(u) + C. Substituting back u=cos⁡2xu = \cos^2 x, we get −tan⁡−1(cos⁡2x)+C-\tan^{-1}(\cos^2 x) + C. Using the identity −tan⁡−1(x)=cot⁡−1(x)−π2-\tan^{-1}(x) = \cot^{-1}(x) - \frac{\pi}{2}, the result can be written as cot⁡−1(cos⁡2x)+C′\cot^{-1}(\cos^2 x) + C', where C′C' is a new arbitrary constant.