Definite Integral of Root Cosine Over Fractional Trigonometric Power

To find the value of the definite integral $I = \int_{0}^{\pi/2} \frac{3\sqrt{\cos\theta}}{\left(\sqrt{\cos\theta} + \sqrt{\sin\theta}\right)^5}\, d\theta$

we can divide both the numerator and the denominator of the integrand by $\cos^{5/2}\theta$:

$I = \int_{0}^{\pi/2} \frac{\dfrac{3\sqrt{\cos\theta}}{\cos^{5/2}\theta}}{\left(\dfrac{\sqrt{\cos\theta} + \sqrt{\sin\theta}}{\sqrt{\cos\theta}}\right)^5}\, d\theta$

Simplifying the numerator and denominator: $\frac{3\sqrt{\cos\theta}}{\cos^{5/2}\theta} = \frac{3}{\cos^2\theta} = 3\sec^2\theta$ $\left(\frac{\sqrt{\cos\theta} + \sqrt{\sin\theta}}{\sqrt{\cos\theta}}\right)^5 = \left(1 + \sqrt{\tan\theta}\right)^5$

Thus, the integral becomes: $I = \int_{0}^{\pi/2} \frac{3\sec^2\theta}{\left(1 + \sqrt{\tan\theta}\right)^5}\, d\theta$

Substitute $u = \tan\theta$, so that $du = \sec^2\theta\, d\theta$.
The limits of integration transform as follows:

• When $\theta = 0$, $u = 0$
• As $\theta \to \frac{\pi}{2}$, $u \to \infty$

The integral now is: $I = \int_{0}^{\infty} \frac{3}{(1 + \sqrt{u})^5}\, du$

Next, substitute $t = \sqrt{u}$, which gives $u = t^2$ and $du = 2t\, dt$:

• When $u = 0$, $t = 0$
• As $u \to \infty$, $t \to \infty$

Substituting these into the integral gives: $I = \int_{0}^{\infty} \frac{3(2t)}{(1+t)^5}\, dt = 6 \int_{0}^{\infty} \frac{t}{(1+t)^5}\, dt$

Now, let $v = 1 + t$, so $t = v - 1$ and $dt = dv$:

• When $t = 0$, $v = 1$
• As $t \to \infty$, $v \to \infty$

Rewriting the integral in terms of $v$: $I = 6 \int_{1}^{\infty} \frac{v - 1}{v^5}\, dv = 6 \int_{1}^{\infty} \left(v^{-4} - v^{-5}\right)\, dv$

Evaluating the integrals: $\int_{1}^{\infty} v^{-4}\, dv = \left[ -\frac{1}{3v^3} \right]_{1}^{\infty} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}$ $\int_{1}^{\infty} v^{-5}\, dv = \left[ -\frac{1}{4v^4} \right]_{1}^{\infty} = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$

Combining the results: $I = 6 \left(\frac{1}{3} - \frac{1}{4}\right) = 6 \left(\frac{1}{12}\right) = 0.5$

Final Answer: The value of the integral is 0.5 (or $\frac{1}{2}$).