Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_0^{\pi /2}\frac{3\sqrt{\cos \theta }}{(\sqrt{\cos \theta }+\sqrt{\sin \theta }{)}^5}d\theta$ equals

Imagine you are standing before a massive, intimidating wall. That is exactly what this integral looks like:

It is complex, it is messy, and it seems to defy standard integration techniques. But in the world of JEE Advanced, the most intimidating problems often hide the most elegant solutions. We are not going to fight this integral head-on; we are going to outsmart it.

The first thing you must notice is the interval: $[0,\pi /2]$. Whenever you see this, your mathematical intuition should scream 'King's Rule!'

The King's Rule is a beautiful property of definite integrals:

It allows us to reflect the function across the midpoint of the interval. Let us apply this to our integral by replacing $\theta$ with $(\pi /2-\theta )$.

Since $\cos (\pi /2-\theta )=\sin \theta$ and $\sin (\pi /2-\theta )=\cos \theta$, our integral transforms into:

Notice how the denominator remained unchanged? That is the magic of symmetry.

Now, we have two expressions for $I$. If we add them together, we get $2I$.

Because the denominators are identical, we simply add the numerators:

Look closely at the numerator. If we factor out the $3$, we get $3(\sqrt{\cos \theta }+\sqrt{\sin \theta })$. This is exactly the base of the denominator!

We can cancel one power, reducing the denominator from power $5$ to power $4$. We are left with:

The beast is already shrinking.

We still have a denominator with a mix of sine and cosine. To solve this, we need to force the expression into a form involving $\tan \theta$ and ${\sec }^2\theta$.

We do this by factoring $\sqrt{\cos \theta }$ out of the denominator. Inside the power of $4$, this becomes $\sqrt{\cos \theta }(1+\sqrt{\tan \theta })$.

When we raise this to the power of $4$, we get ${\cos }^2\theta (1+\sqrt{\tan \theta }{)}^4$. Since $1/{\cos }^2\theta ={\sec }^2\theta$, we can move the cosine term to the numerator.

Our integral becomes:

This is perfect! We have $\tan \theta$ in the denominator and its derivative, ${\sec }^2\theta$, in the numerator.

Let $u=\tan \theta$. Then $du={\sec }^2\theta d\theta$. As $\theta$ goes from $0$ to $\pi /2$, $u$ goes from $0$ to $\infty$.

Our integral is now:

To simplify further, let $t=1+\sqrt{u}$. This implies $\sqrt{u}=t-1$, so $u=(t-1{)}^2$, and $du=2(t-1)dt$.

Substituting these, we get:

Splitting this into $6{\int }_1^{\infty }(t^{-3}-t^{-4})dt$, we are left with a simple polynomial integration.

Evaluating this from $1$ to $\infty$ gives us $6(0-(-1/6))=1$. Thus, $2I=1$, which means $I=0.5$. We have tamed the beast!