$\int \frac{5}{\sin ^{3} x} d x=-\frac{5 \cos x}{2 \sin ^{2} x}-\frac{5}{4} \ln \left|\frac{\cos x-1}{\cos x+1}\right|+C$

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The Answer of $\int \frac{5}{\sin ^{3} x} d x=-\frac{5 \cos x}{2 \sin ^{2} x}-\frac{5}{4} \ln \left|\frac{\cos x-1}{\cos x+1}\right|+C$

∫5sin⁡3xdx=−5cos⁡x2sin⁡2x−54ln⁡∣cos⁡x−1cos⁡x+1∣+C\int \frac{5}{\sin ^{3} x} d x=-\frac{5 \cos x}{2 \sin ^{2} x}-\frac{5}{4} \ln \left|\frac{\cos x-1}{\cos x+1}\right|+C∫sin3x5​dx=−2sin2x5cosx​−45​ln​cosx+1cosx−1​​+C