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Use integration by parts to find a formula for the integral of sinnx

Using integration by parts, prove the validity of the formula:

    \[ \int \sin^n x \, dx = -\sin^{n-1} x  \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx. \]

Then, using the pythagorean identity, \cos^2 x = 1 - \sin^2 x, establish the formula:

    \[ \int \sin^n x \, dx = - \frac{\sin^{n-1} x  \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx. \]


Proof. To apply integration by parts let

    \begin{align*}  u &= \sin^{n-1} x & \implies && du &= (n-1) \sin^{n-2} x \cos x \, dx \\  dv &= \sin x \, dx & \implies && v &= -\cos x. \end{align*}

Then, applying integration by parts we have

    \begin{align*}  \int \sin^n x \, dx &= \int u \, dv \\  &= uv - \int v \, du \\  &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx. \end{align*}

Then, we use the pythagorean identity, \cos^2 x = 1 - \sin^2 x and obtain

    \begin{align*}  &&\int \sin^n x \, dx &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \, (1 - \sin^2 x) \, dx \\  \implies && \int \sin^n x \, dx &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \, dx - (n-1) \int \sin^n x \, dx \\  \implies && n \int \sin^n x \, dx &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \, dx \\  \implies && \int \sin^n x \, dx &= -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx. \qquad \blacksquare \end{align*}

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