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Calculus · real student question

Evaluate the integral of sin(x)/sin^2(4 cos(x)) dx.

Question

Evaluate ∫sin⁡xsin⁡2(4cos⁡x) dx.\int \frac{\sin x}{\sin^{2}(4\cos x)}\,dx.

Step-by-step solution

  1. Spot the inner function and its derivative. The awkward part is the composite sin⁡2(4cos⁡x)\sin^{2}(4\cos x), whose inner function is 4cos⁡x4\cos x. Its derivative is −4sin⁡x-4\sin x, and a factor sin⁡x\sin x is already sitting in the numerator, so the substitution u=4cos⁡xu=4\cos x will clear the integral completely.

  2. Carry out the substitution. From u=4cos⁡xu=4\cos x we get du=−4sin⁡x dxdu=-4\sin x\,dx, hence sin⁡x dx=−14 du\sin x\,dx=-\frac14\,du, and ∫sin⁡xsin⁡2(4cos⁡x) dx=−14∫dusin⁡2u=−14∫csc⁡2u du.\int\frac{\sin x}{\sin^{2}(4\cos x)}\,dx=-\frac14\int\frac{du}{\sin^{2}u}=-\frac14\int\csc^{2}u\,du .

  3. Use the standard cosecant-squared antiderivative. Because dducot⁡u=−csc⁡2u\frac{d}{du}\cot u=-\csc^{2}u, we have ∫csc⁡2u du=−cot⁡u+C\int\csc^{2}u\,du=-\cot u+C, so −14∫csc⁡2u du=14cot⁡u+C.-\frac14\int\csc^{2}u\,du=\frac14\cot u+C .

  4. Return to the original variable. Substituting u=4cos⁡xu=4\cos x back gives ∫sin⁡xsin⁡2(4cos⁡x) dx=14cot⁡(4cos⁡x)+C.\int\frac{\sin x}{\sin^{2}(4\cos x)}\,dx=\frac14\cot(4\cos x)+C .

  5. Verify by differentiating. ddx[14cot⁡(4cos⁡x)]=14(−csc⁡2(4cos⁡x))⋅(−4sin⁡x)=sin⁡xsin⁡2(4cos⁡x)\frac{d}{dx}\left[\frac14\cot(4\cos x)\right]=\frac14\left(-\csc^{2}(4\cos x)\right)\cdot(-4\sin x)=\frac{\sin x}{\sin^{2}(4\cos x)}, which is the original integrand, so the antiderivative is correct on every interval where sin⁡(4cos⁡x)≠0\sin(4\cos x)\ne 0.

Answer

14cot⁡(4cos⁡x)+C\frac{1}{4}\cot(4\cos x)+C

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