If int frac{1}{cos^3 x sqrt{sin 2x}} dx = p(t | vedclass

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Question

(C) Given integral $I = \int \frac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx = \int \frac{1}{\cos^3 x \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}} dx = \frac{1}

Vedclass · Accepted Answer

$p = \frac{\sqrt{2}}{5}, q = 5$

Vedclass · Answer

(C) Given integral $I = \int \frac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx = \int \frac{1}{\cos^3 x \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}} dx = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^{3.5} x \sqrt{\sin x}} dx$.Divide numerator and denominator by $\cos^2 x$: $I = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{	an x}} dx$.Let $	an x = t$, then $\sec^2 x dx = dt$. So, $I = \frac{1}{\sqrt{2}} \int t^{-1/2} dt = \frac{1}{\sqrt{2}} \cdot \frac{t^{1/2}}{1/2} = \sqrt{2} \sqrt{	an x} + c$.Wait, the form is $p(	an^2 x + q)\sqrt{	an x}$. Let's rewrite $\sqrt{2} \sqrt{	an x}$ as $\frac{\sqrt{2}}{5} (5 	an x) \sqrt{	an x}$. This does not match. Let's re-evaluate: $\int \frac{\sec^2 x}{\sqrt{	an x}} dx = 2\sqrt{	an x}$.Actually, $\int \frac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx = \int \frac{1}{\sqrt{2} \cos^3 x \sqrt{	an x} \sqrt{\cos x}} dx = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{	an x}} dx = \frac{1}{\sqrt{2}} \int \frac{(1+	an^2 x) \sec^2 x}{\sqrt{	an x}} dx$.Let $	an x = t$, $dt = \sec^2 x dx$. $I = \frac{1}{\sqrt{2}} \int (1+t^2) t^{-1/2} dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) dt = \frac{1}{\sqrt{2}} [2t^{1/2} + \frac{2}{5} t^{5/2}] = \sqrt{2} \sqrt{	an x} + \frac{\sqrt{2}}{5} 	an^2 x \sqrt{	an x} = \frac{\sqrt{2}}{5} \sqrt{	an x} (5 + 	an^2 x)$.Comparing with $p(	an^2 x + q)\sqrt{	an x}$, we get $p = \frac{\sqrt{2}}{5}$ and $q = 5$.

If $\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} dx = p(\tan^2 x + q)\sqrt{\tan x} + c$, then the values of $p$ and $q$ respectively are

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