1.

1 - cos x=

Answer» we know \( \sin^2 x=\frac{1-\cos 2x}{2} \). so we have:

\( \int \sin^2 x \; \mathrm{d}x=\int \frac{1-\cos 2x}{2} \; \mathrm{d}x=\int \frac{1}{2} \; \mathrm{d}x-\int \frac{\cos 2x}{2} \; \mathrm{d}x \)

\( \int \sin^2 x \; \mathrm{d}x=\frac{1}{2}x-\frac{1}{4}\sin 2x+c \)


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