Question

The value of integral $\int \frac{d\theta }{co{s}^3\theta \sqrt{sin2\theta }}$ can be expressed as irrational function of $tan\theta$ as

• A. $\frac{\sqrt{2}}{5}\left(\sqrt{ta{n}^2\theta +5}\right)tan\theta +c$
• B. $\frac{2}{5}(ta{n}^2\theta +5)\sqrt{tan\theta }+c$
• C. $\frac{\sqrt{2}}{5}(ta{n}^2\theta +5)\sqrt{tan\theta }+c$
• D. $\sqrt{\frac{2}{5}}(ta{n}^2\theta +5)\sqrt{tan\theta }+c$

Answer 1

The correct option is C $\frac{\sqrt{2}}{5}(ta{n}^2\theta +5)\sqrt{tan\theta }+c$

$\int \frac{d\theta }{co{s}^3\theta \sqrt{sin2\theta }}=\int \frac{d\theta }{co{s}^4\theta \sqrt{2tan\theta }}$

$=\int \frac{se{c}^2\theta (1+ta{n}^2\theta )}{\sqrt{2tan\theta }}d\theta$

$=\frac{1}{\sqrt{2}}\int \frac{1+t^2}{\sqrt{t}}dt$ [Substitute

$tan\theta =t$]

$=\frac{1}{\sqrt{2}}(2\sqrt{t}+\frac{t^{5/2}}{5/2})=\frac{\sqrt{2}}{5}\sqrt{tan\theta }(5+ta{n}^2\theta )+c$