AP Calculus AB and BC: Chapter 6 – Techniques of Integration : 6.2 – Trigonometric Integral Study Notes

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Trigonometric Integrals

Trigonometric Identities
$
\begin{aligned}
& \sin ^2 x+\cos ^2 x=1 \quad \tan ^2 x+1=\sec ^2 x \quad \cot ^2 x+1=\csc ^2 x \\
& \sin ^2 x=\frac{1-\cos 2 x}{2} \quad \cos ^2 x=\frac{1+\cos 2 x}{2} \\
& \sin 2 x=2 \sin x \cos x
\end{aligned}
$

Guidelines for Evaluating $\int \sin ^m x \cos ^n x d x$.
1. If $m$ is odd, save one sine factor and use $\sin ^2 x=1-\cos ^2 x$ to express the remaining factor in terms of cosine.

Example 1

  • Evaluate $\int \sin ^3 x \cos ^2 x d x$
▶️Answer/Explanation

Solution
$\begin{array}{lll}=\int \sin ^2 x \cos ^2 x(\sin x) d x & & \text { One sine factor is saved. } \\ =\int\left(1-\cos ^2 x\right) \cos ^2 x(\sin x) d x & & \sin ^2 x=1-\cos ^2 x \\ =\int\left(\cos ^2 x-\cos ^4 x\right) \sin x d x & & \text { Multiply. } \\ =\int\left(u^2-u^4\right)(-d u) & & u=\cos x, d u=-\sin x d x \\ =-\frac{1}{3} \cos ^3 x+\frac{1}{5} \cos ^5 x+C & \end{array}$

2. If $n$ is odd, save one cosine factor and use $\cos ^2 x=1-\sin ^2 x$ to express the remaining factor in terms of sine.

Example 2

  • Evaluate $\int \cos ^5 x d x$
▶️Answer/Explanation

Solution 

$\int \cos ^5 x d x$

$=\int \cos ^4 x(\cos x) d x$                                                     One cosine factor is saved.

$=\int\left(1-\sin ^2 x\right)^2 \cos x d x                                  \quad \cos ^2 x=1-\sin ^2 x$

$=\int\left(1-u^2\right)^2 d u \quad u=\sin x, d u=\cos x d x$

$=\int\left(1-2 u^2+u^4\right) d u$                                                     Multiply.

$\begin{aligned} & =u-\frac{2}{3} u^3+\frac{1}{5} u^5+C \\ & =\sin x-\frac{2}{3} \sin ^3 x+\frac{1}{5} \sin ^5 x+C\end{aligned}$

Example3

  • If both $m$ and $n$ are even, substitute $\sin ^2 x=\frac{1-\cos 2 x}{2}$ and $\cos ^2 x=\frac{1+\cos 2 x}{2}$ to reduce the integrand to lower powers of $\cos 2 x$.

Evaluate $\int \sin ^2 x \cos ^2 x d x$.

▶️Answer/Explanation

Solution
$
\begin{aligned}
& \int \sin ^2 x \cos ^2 x d x \\
& =\int\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1+\cos 2 x}{2}\right) d x \quad \sin ^2 x=\frac{1-\cos 2 x}{2}, \cos ^2 x=\frac{1+\cos 2 x}{2} \\
& =\frac{1}{4} \int\left(1-\cos ^2 2 x\right) d x \\
& =\frac{1}{4} \int\left(1-\frac{1+\cos 4 x}{2}\right) d x \quad \cos ^2 2 x=\frac{1+\cos 4 x}{2} \\
& =\frac{1}{8} \int(1-\cos 4 x) d x \\
& =\frac{1}{8} x-\frac{1}{32} \sin 4 x+C \\
&
\end{aligned}
$

Example5

  • $
    \int \sin ^3 n x d x=
    $

(A) $\frac{1}{3 n} \sin ^3 n x-\frac{1}{n} \sin n x+C$

(B) $\frac{1}{3 n} \cos ^3 n x-\frac{1}{n} \cos n x+C$

(C) $\frac{1}{3 n} \sin ^3 n x-\frac{1}{n} \sin n x+C$

(D) $\frac{1}{3 n} \cos ^3 n x-\frac{1}{n} \sin n x+C$

▶️Answer/Explanation

Ans:C

Example6

  • $
    \int \cos ^3 x \sqrt{\sin x} d x=
    $

(A) $\frac{1}{3}(\cos x)^3-\frac{2}{5}(\cos x)^{5 / 2}+C$

(B) $\frac{2}{3}(\cos x)^{3 / 2}-\frac{2}{7}(\sin x)^{7 / 2}+C$

(C) $\frac{2}{3}(\sin x)^{3 / 2}-\frac{2}{7}(\sin x)^{7 / 2}+C$

(D) $\frac{2}{3}(\sin x)^{3 / 2}-\frac{2}{5}(\cos x)^{5 / 2}+C$

▶️Answer/Explanation

Ans:B

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