Integrating sin² x, cos² x and tan² x (with Examples and Practice Questions)
If you want to integrate sin² x, cos² x or tan² x, you can perform integration by substitution.
An alternative (and perhaps easier method) is to use trigonometric identities to rewrite the integral in terms of sin x, cos x, or sec² x, all of which have easy integrals. This makes it much easier to integrate.
Note: If you don't know how to integrate sin x, cos x or sec² x, then you'll need to learn those first.
Rewriting sin² x, cos² x and tan² x
From the double angle formulae we know that:
\( \cos 2x \equiv 1 - 2\sin^2 x \) and
\( \cos 2x \equiv 2\cos^2 x - 1 \)
We also have the following trigonometric identity including tan x:
\( 1 + \tan^2 x \equiv \sec^2 x \)
We can rearrange all three equations to get the following identities involving sin² x, cos² x and tan² x:
\( \sin^2 x \equiv \frac{1}{2} - \frac{1}{2}\cos2x \)
\( \cos^2 x \equiv \frac{1}{2} + \frac{1}{2}\cos2x \)
\( \tan^2 x \equiv \sec^2 x - 1 \)
We can then substitute these identities into integrands involving sin² x, cos² x or tan² x in order to rewrite them before integrating.
Integrating sin² x, cos² x and tan² x: Examples
Example 1:
Find \( \int 6\cos^2 x \, dx \).
Solution:
Substitute the identity \( \cos^2 x \equiv \frac{1}{2}(1 + \cos 2x) \) into the integrand.
\[ \int 6\sin^2 x \, dx = \int 6 \cdot \frac{1}{2}(1 + \cos 2x) \, dx. \]
Simplify by taking out the factor of 3:
\[ = 3 \int (1 + \cos 2x) \, dx. \]
Now, integrate each term separately:
\[ = 3 \left[ x + \frac{\sin 2x}{2} \right] + C. \]
So, the value of the integral is:
\[ \int 6\sin^2 x \, dx = 3x + \frac{3}{2} \sin 2x + C. \]
Example 2:
Find the value of \( \int_{0}^{\pi} 10\sin^2 2x \, dx \).
Solution:
Use the identity \( \sin^2 x \equiv \frac{1}{2}(1 - \cos 2x) \) to rewrite \( \sin^2 2x \):
\[ \sin^2 2x = \frac{1}{2}(1 - \cos 4x). \]
Substitute this into the integral:
\[ \int_{0}^{\pi} 10\sin^2 2x \, dx = \int_{0}^{\pi} 10 \cdot \frac{1}{2}(1 - \cos 4x) \, dx. \]
Take out the factor of 5:
\[ = 5 \int_{0}^{\pi} (1 - \cos 4x) \, dx. \]
Now, integrate each term and substitute the values of \( \pi \) and 0:
\[ = 5 \left[ x - \frac{\sin 4x}{4} \right]_{0}^{\pi} \]
\[ = 5 \left[ \left(\pi - \frac{\sin 4\pi}{4} \right) - \left( 0 - \frac{\sin 0}{4} \right) \right] \]
Now simplify:
\[ = 5 \left[ \left(\pi - 0 \right) - \left( 0 - 0 \right) \right] = 5\pi \]
So, the value of the integral is:
\[ \int_{0}^{\pi} 10\sin^2 2x \, dx = 5\pi. \]
Practice Questions
Question 1
Find \( \int 6\tan^2 (3x) \, dx \).