If $\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}=( an x)^{\mathrm{A}}+\mathrm{C}( an x)^{\mathrm{B}}+\mathrm{K}$, where K is a constant of integration, then the value of $5(\mathrm{~A}+\mathrm{B}+\mathrm{C})$ is equal to [Note: The question has been modified to get the correct answer.]

Question

If $\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}=(	an x)^{\mathrm{A}}+\mathrm{C}(	an x)^{\mathrm{B}}+\mathrm{K}$, where K is a constant of integration, then the value of $5(\mathrm{~A}+\mathrm{B}+\mathrm{C})$ is equal to [Note: The question has been modified to get the correct answer.]

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}} \ & 	ext { Put } \sin 2 x=\frac{2 	an x}{1+	an ^2 x} \ & 	herefore \quad \mathrm{I}=\int \frac{\sec ^2 x \cdot \sec ^2 x}{2 \sqrt{x 	an x}} \mathrm{~d} x \end{aligned}$ Let $	an x=\mathrm{t}$ $	herefore \quad \sec ^2 \mathrm{~d} x=\mathrm{dt}$ $\begin{aligned} 	herefore \quad I & =\frac{1}{2} \int \frac{1+t^2}{\sqrt{t}} d t=\frac{1}{2} \int\left(\frac{1}{\sqrt{t}}+t^{\frac{3}{2}}ight) d t \ & =\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}ight]+k=(	an 	heta)^{\frac{1}{2}}+\frac{1}{5}(	an 	heta)^{\frac{5}{2}}+k\end{aligned}$ Comparing with $(	an x)^{\mathrm{A}}+\mathrm{C}(	an x)^{\mathrm{B}}+\mathrm{k}$, we get $\begin{aligned} & A=\frac{1}{2}, B=\frac{5}{2}, C=\frac{1}{5} \ 	herefore \quad & 5(A+B+C)=5\left(\frac{1}{2}+\frac{5}{2}+\frac{1}{5}ight)=16 \end{aligned}$ [Note: In the question, $\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{x \sin 2 x}}$ is changed to $\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}$ to apply appropriate textual concepts.]

If $\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}=(\tan x)^{\mathrm{A}}+\mathrm{C}(\tan…

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