Integrating Trigonometric Functions by Double Angle Formula

Integrating Trigonometric Functions can be done by Double Angle Formula reducing the power of trigonometric functions.

\( \begin{aligned} \displaystyle
\cos{2A} &= 2\cos^2{A}-1 \\
&= 1-2\sin^2{A} \\
&= \cos^2{A}-\sin^2{A}
\end{aligned} \)

Practice Questions

Question 1

Find \( \displaystyle \int{\cos^2{x}}dx \).

\( \begin{aligned} \displaystyle
2\cos^2{x}-1 &= \cos{2x} \\
2\cos^2{x} &= \cos{2x} + 1 \\
\cos^2{x} &= \frac{1}{2}(\cos{2x}+1) \\
\int{\cos^2{x}}dx &= \frac{1}{2} \int{(\cos{2x}+1)}dx \\
&= \frac{1}{2}\Big(\frac{1}{2}\sin{2x}+x\Big) + C \\
&= \frac{1}{4} \sin{2x} + \frac{1}{2}x + C
\end{aligned} \)

Question 2

Find \( \displaystyle \int{\sin^2{x}}dx \).

\( \begin{aligned} \displaystyle
\cos{2x} &= 1-2\sin^2{x} \\
2\sin^2{x} &= 1-\cos{2x} \\
\sin^2{x} &= \frac{1}{2}(1-\cos{2x}) \\
\int{\sin^2{x}}dx &= \frac{1}{2} \int{(1-\cos{2x})}dx \\
&= \frac{1}{2}\Big(x-\frac{1}{2}\sin{2x}\Big) + C \\
&= \frac{1}{2}x-\frac{1}{4}\sin{2x} + C
\end{aligned} \)

Question 3

Find \( \displaystyle \int{\sin^4{x}}dx \).

\( \begin{aligned} \displaystyle
\sin^4{x} &= (\sin^2{x})^2 \\
&= \frac{1}{4}(1-\cos{2x})^2 \\
&= \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}) \\
&= \frac{1}{4}\Big[1-2\cos{2x}+\frac{1}{2}(1+\cos{4x})\Big] \\
&= \frac{1}{4}\Big[\frac{3}{2}-2\cos{2x}+\frac{1}{2}\cos{4x}\Big] \\
&= \frac{3}{8}-\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x} \\
\int{\sin^4{x}}dx &= \int{\Big(\frac{3}{8}-\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}\Big)}dx \\
&= \frac{3}{8}x-\frac{1}{4}\sin{2x} + \frac{1}{32}\sin{4x} + C
\end{aligned} \)

Question 4

Find \( \displaystyle \int{(\cos^2x-\sin^2x)}dx \).

\( \begin{aligned} \displaystyle
\int{(\cos^2x-\sin^2x)}dx &= \int{\cos{2x}}dx \\
&= \frac{1}{2}\sin{2x} + C
\end{aligned} \)

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