Integration of cos inverse x

Integration of cos inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it

\[ \int \cos^{-1}x \, dx = x \cos^{-1}x – \sqrt{1 – x^2} + C \]

where (C) is the constant of integration.

Proof of integration of cos inverse x

We can prove it using integration by parts and integration by substitution

Integration by parts

To find the integral we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx $
In our case, we can let $ f(x) = \cos^{-1}x $ and $ g(x) = 1 $. Then

• $ \frac {df(x)}{dx} = -\frac{1}{\sqrt {1-x^2}} \, dx $
• $\int g(x) dx = \int dx = x $

Now, substitute these into the integration by parts formula:

$$
\int \cos^{-1}x \, dx = x \cos^{-1}x – \int x \cdot \frac{-1}{\sqrt {1-x^2}} \, dx
$$

$ =x \cos^{-1}x – \int \frac{x}{\sqrt {1-x^2}} \, dx \\
=x \cos^{-1}x – \frac {1}{2} \int \frac{-2x}{\sqrt {1-x^2}} \, dx $

Now lets calculate the second integral separately $\int \frac{-2x}{\sqrt {1-x^2}} \, dx $

Let $t= 1-x^2$
then $dt=-2x dx$
Therefore

$\int \frac{-2x}{\sqrt {1-x^2}} \, dx = \int \frac{1}{\sqrt {t}} \, dx \\
= \frac {\sqrt {t}}{1/2} = 2 \sqrt {1-x^2}$

Substituting this value in main integral , we get

$$
\int \cos^{-1}x \, dx = x \cos^{-1}x – \sqrt {1-x^2} + C
$$

Integration by substitution

let $\theta = \cos^{-1}x$
$ cos \theta = x$
$-sin \theta d \theta = dx$

$$
\int \cos^{-1}x \, dx = -\int \theta sin \theta d \theta
$$

Now solving this using integration parts

$ = -\theta \int sin \theta d \theta + \int (\frac {d}{d \theta } \theta ) .(\int sin \theta d \theta) \; \theta \\
= \theta \cos \theta – \int \cos \theta \; d \theta \\
=\theta \cos \theta – \sin \theta $

Substituting back the values

$\int \cos^{-1}x \, dx = x \cos^{-1}x – \sqrt {1-x^2} + C$

Definite Integral of cos inverse x

To find the definite integral of $\cos^{-1}x$ 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 $\cos^{-1}x$ from $a$ to $b$ is given by:

$$\int_{a}^{b} \cos^{-1}x \, dx = b \cos^{-1}b – a \cos^{-1}a + \sqrt{1 – a^2} – \sqrt{1 – b^2} $$

This expression represents the accumulated area under the curve of $\cos^{-1}x$ from $x = a$ to $x = b$.

Solved Examples on Integration of cos inverse x

Question 1

$$\int_{0}^{1} \cos^{-1}x \, dx$$

Solution

$\int_{0}^{1} \cos^{-1}x \, dx = [x \cos^{-1}x – \sqrt{1 – x^2}] _0 ^1 \\
= 1. cos^{-1} 1 – 0 – 0 +1= 1$

Question 2

integral of $x \cos^{-1}x $

Solution

Applying integration by parts
$\int u \, dv = uv – \int v \, du$.
Let $u = \cos^{-1}x$. Then, $du = \frac{-dx}{\sqrt{1 – x^2}}$.
Let $dv = x \, dx$. Then, $v = \frac{x^2}{2}$.
Then
$ \int x \cos^{-1}x \, dx = \frac{x^2}{2} \cos^{-1}x – \int \frac{x^2}{2} \cdot \frac{-dx}{\sqrt{1 – x^2}} \
=\frac{x^2}{2} \cos^{-1}x – \frac {1}{2} \int \frac{-x^2}{\sqrt{1 – x^2}} \; dx$
Now
$\int \frac{-x^2}{\sqrt{1 – x^2}} \; dx = \int \frac{1-x^2 – 1}{\sqrt{1 – x^2}} \; dx \
= \int \sqrt {1-x^2} dx – \int \frac {1}{\sqrt {1-x^2}} dx $

Now $\int \frac {1}{\sqrt {1-x^2}} dx = \sin^{-1} x $
Also $ \int \sqrt{1 – x^2} \, dx $ can be found as below
Here, we can use the trigonometric substitution:
$ x = \sin(\theta) $
$ dx = \cos(\theta) \, d\theta $
Therefore
$ \int \sqrt{1 – \sin^2(\theta)} \cos(\theta) \, d\theta $
$ = \int \cos^2(\theta) \, d\theta $
This can be integrated using the half-angle formula:
$ \int \frac{1 + \cos(2\theta)}{2} \, d\theta $
$ = \frac{\theta}{2} + \frac{\sin(2\theta)}{4} + C $
Now, reverting the substitution:
$ \sin^{-1}(x) = \theta $
$ 2\sin(\theta)\cos(\theta) = \sqrt{1-x^2} $
Therefore
$ = \frac{\sin^{-1}(x)}{2} + \frac{x \sqrt{1 – x^2}}{2} + C $

Applying these steps to the integral of $x \cos^{-1}x$, we obtained:

$$ \frac{x^2 \cos^{-1}x}{2} – \frac{x \sqrt{1 – x^2}}{4} + \frac{3 \sin^{-1}x}{4} + C $$

where $C$ is the constant of integration.

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