Trigonometric identities can be used to simplify and evaluate integrals. In some cases trigonometric identities may be used to write the integrand in a form which can be integrated directly. In others, a trigonometric substitution can transform the integral into a simpler form.
Find $\displaystyle{ \int\sin^2{x}\mathrm{d} x. }$
Recall the trigonometric identity $\cos{2\theta}=1-2\sin^2{\theta}$.
Rearranging this gives $\sin^2{\theta}=\dfrac{1}{2}(1-\cos{2\theta})$, and the integral can therefore be written as:
\[\int\sin^2{x}\mathrm{d} x = \int\dfrac{1}{2}(1-\cos{2x})\mathrm{d} x.\]
The integral is now in a form which can be evaluated directly:
\begin{align} \int\dfrac{1}{2}(1-\cos{2x})\mathrm{d} x &= \dfrac{1}{2} \int \left( 1 - \cos{2x}\right) \mathrm{d} x \\ &= \dfrac{1}{2} \left( x - \dfrac{1}{2}\sin{2x} \right) +C \\ &= \frac{x}{2} - \frac{1}{4}\sin 2x + C. \end{align}
Find $\displaystyle{ \int\cos^3(x)\sin^2(x)\mathrm{d} x. }$
Here there is no obvious identity which simplifies the integral. It is necessary to manipulate the integrand in such a way that an identity can be used.
Write $\cos^3{x}=\cos{x}\cos^2{x}$. Then by the identity $\cos^2{\theta}+\sin^2{\theta}=1$, we have $\cos^2{\theta}=1-\sin^2{\theta}$ and $\cos^3{x}$ can be written in the form:
\[\cos^3{x}=\cos{x}\bigl(1-\sin^2{x}\bigl).\]
Substituting this expression into the integral gives:
\[\int\cos^3{x}\sin^2{x}\mathrm{d} x=\int\cos{x}\bigl(1-\sin^2{x}\bigl)\sin^2{x}\mathrm{d} x.\]
Now introduce the substitution $u=\sin{x}$. Then $\mathrm{d} u =\dfrac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x=\cos{x}\mathrm{d} x$.
Substitute $u=\sin{x}$ and $\mathrm{d} u=\cos{x}\mathrm{d} x$ into the integral: \begin{align} \int \cos{x} (1-\sin^2 x)\sin^2 x \mathrm{d} x &= \int (1-\sin^2 x ) \cdot \sin^2 x \cdot \cos x \mathrm{d} x \\ &= \int(1-u^2) u^2 \mathrm{d} u. \end{align}
This can be evaluated directly in terms of $u$: \begin{align} \int (1 - u^2) u^2 \mathrm{d} u &=\int u^2-u^4\mathrm{d} u \\ &=\dfrac{u^3}{3}-\dfrac{u^5}{5}+C. \end{align}
Substituting $u=\sin{x}$ into this expression gives the value of the integral in terms of the original variable $x$:
\[\int\cos^3{x}\sin^2{x}\mathrm{d} x=\dfrac{1}{3}\sin^3{x}-\dfrac{1}{5}\sin^5{x}+C.\]
Find $\displaystyle{ \int\dfrac{1}{1+x^2}\mathrm{d} x }$ by substituting $x=\tan{\theta}$.
If $x=\tan{\theta}$ then \begin{align} \mathrm{d} x &= \dfrac{\mathrm{d} x}{\mathrm{d} \theta}\mathrm{d} \theta \\ &= \dfrac{\mathrm{d
{\mathrm{d} \theta}\bigl[\tan{\theta}\bigl]\mathrm{d} \theta \\ &= \sec^2{\theta}\mathrm{d} \theta. \end{align}
Substituting these expressions into the integral gives:
\[\int\dfrac{1}{1+x^2}\mathrm{d} x=\int\dfrac{\sec^2{\theta} }{1+\tan^2{\theta} }\mathrm{d} \theta.\]
Recall the trigonometric identity $\sec^2{\theta}=1+\tan^2{\theta}$. Making use of this identity allows the integral to be evaluated directly:
\begin{align} \int\dfrac{\sec^2{\theta} }{1+\tan^2{\theta} }\mathrm{d} \theta &= \int\dfrac{\sec^2{\theta} }{\sec^2{\theta} }\mathrm{d} \theta \\ &= \int 1 \mathrm{d} \theta \\ &= \theta + C. \end{align}
To transform this result back into an expression in the original variable $x$, note that $\theta=\arctan{x}$.
Hence,
\[\int\dfrac{1}{1+x^2}\mathrm{d} x=\arctan{x}+C.\]
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Prof. Robin Johnson uses trigonometric identities to find $\begin{align}\int \sin^3{x}\;\mathrm{d}x\end{align}$.
Prof. Robin Johnson uses a trigonometric identity to find the integral of a trigonometric function raised to a power, $\begin{align}\int\cos^3{(2-5x)}\;\mathrm{d}x\end{align}$.
Prof. Robin Johnson uses a trigonometric identity to find $\begin{align}\int_0^1\dfrac{\mathrm{d}x}{\sqrt{1-x^2} }\end{align}$ by subsititution.