integration of sin x cos x dx

Integration of sin x cos x dx can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of sin x cos x dx are

I

\[
\int \sin(x) \cos (x)\, dx = -\frac {1}{4} \cos (2x) + C
\]

II

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

III

\[
\int \sin(x) \cos (x)\, dx = \frac {1}{2} \sin^2 (x) + C
\]

Lets check out the proof of each of these integration of sin x cos x dx

Proof of I

Integration of cos x sin x can be solved by using a clever trick involving multiplying and dividing by 2. Here’s how it’s done:

$ \int \sin(x) \cos (x)\, dx = \frac {1}{2} \int 2\sin(x) \cos (x)\, dx = \frac {1}{2} \int \sin (2x) \, dx$

Now taking 2x= t
$2 dx= dt$
$dx= \frac {dt}{2}$

$ \int \sin(x) \cos (x)\, dx = \frac {1}{4} \int \sin (t) \; dt = -\frac {1}{4} \cos (2x) + C$

where (C) is the constant of integration. This is the integral of $(\sin(x) \cos(x))$.

Proof of II

$ \int \sin(x) \cos (x)\, dx$

let $\cos x = t$
$-\sin x dx = dt$
substituting these

$ \int \sin(x) \cos (x)\, dx = – \int t \; dt = -\frac {t^2}{2} + C$

Hence

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

Proof of III

$ \int \sin(x) \cos (x)\, dx$

let $\sin x = t$
$\cos x dx = dt$
substituting these

$ \int \sin(x) \cos (x)\, dx = \int t \; dt = \frac {t^2}{2} + C$

Hence

\[
\int \sin(x) \cos (x)\, dx = \frac {1}{2} \sin^2 (x) + C
\]

Definite Integral of sin x cosx dx

To find the definite integral of $\sin x \cos x $ over a specific interval, we use the same approach as with the indefinite integral, but we’ll apply the limits of integration at the end.

The definite integral of $\sin x \cos x $ from $a$ to $b$ is given by:

$$\int_{a}^{b} \sin(x) \cos (x) \, dx = – \frac {1}{4}\cos (2b) + \frac {1}{4} \cos (2a) $$

This expression represents the accumulated area under the curve of $\sin x \cos x $ from $x = a$ to $x = b$

Solved Examples on integration of sin x cos x dx

Question 1

$\int_{0}^{\pi} \sin(x) \cos (x) \, dx $

Solution

$\int_{0}^{\pi} \sin(x) \cos (x) \, dx =- \frac {1}{4}\cos (2\pi) + \frac {1}{4} \cos (0)= 0$

Question 2

Evaluate the definite integral of $\cos x \sin x$ from $0$ to $\pi/4$ and then use the result to find the area enclosed between the curve $y = \cos x \sin x$, the x-axis, and the lines $x = 0$ and $x = \pi/4$.

Solution
$$
\int_{0}^{\pi/4} \cos x \sin x \, dx = \frac{1}{2} \int_{0}^{\pi/4} \sin(2x) \, dx
$$

We can calculate this integral:

$$
\frac{1}{2} \left[ -\frac{1}{2} \cos(2x) \right]_{0}^{\pi/4}
$$

$$
= \frac{1}{2} \left( -\frac{1}{2} \cos(\pi/2) + \frac{1}{2} \cos(0) \right)
$$

$$
= \frac{1}{2} \left( -\frac{1}{2} \cdot 0 + \frac{1}{2} \cdot 1 \right)
= \frac{1}{4}
$$

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