Question

$\int \sqrt{\frac{1+\sin \theta -{\sin }^2\theta -{\sin }^3\theta }{2\sin \theta -1}}d\theta$ is equal to
[Note: Where C is constant of integration.]

• A. $\frac{1}{2}\sqrt{\sin \theta -\cos 2\theta }+\frac{3}{4\sqrt{2}}{\log }_e(4\sin \theta +1+2\sqrt{2}\sqrt{\sin \theta -\cos 2\theta })+C$
• B. $\frac{1}{2}\sqrt{\sin \theta +\cos 2\theta }+\frac{3}{4\sqrt{2}}{\log }_e(4\sin \theta -1+2\sqrt{2}\sqrt{\sin \theta +\cos 2\theta })+C$
• C. $\frac{1}{2\sqrt{2}}\sqrt{\sin \theta -\cos 2\theta }+\frac{3}{4}{\log }_e(4\sin \theta +1-\sqrt{\sin \theta -\cos 2\theta })+C$
• D. $\frac{1}{2}\sqrt{\sin \theta +\cos 2\theta }+\frac{3}{4\sqrt{2}}{\log }_e(4\sin \theta +1-\sqrt{\sin \theta -\cos 2\theta })+C$

Answer 1

Answer: (A)

$\sqrt{1+\sin \theta -{\sin }^2\theta -{\sin }^3\theta }=\sqrt{(1+\sin \theta )\left(1-{\sin }^2\theta \right)}=\sqrt{1+\sin \theta }\cos \theta$
With $x=\sin \theta$, the given integral is
$\int \sqrt{\frac{1+x}{2x-1}}dx=\int \frac{1+x}{\sqrt{2x^2+x-1}}dx=\frac{1}{4}\int \frac{4x+1}{\sqrt{2x^2+x-1}}dx+\frac{3}{4\sqrt{2}}\int \frac{dx}{\sqrt{{\left(x+\frac{1}{4}\right)}^2-\frac{9}{16}}}$
$=\frac{1}{2}\sqrt{2x^2+x-1}+\frac{3}{4\sqrt{2}}\log \left(x+\frac{1}{4}+\sqrt{{\left(x+\frac{1}{4}\right)}^2-\frac{9}{16}}\right)+C$
$=\frac{1}{2}\sqrt{2x^2+x-1}+\frac{3}{4\sqrt{2}}\log \left(4x+1+2\sqrt{2}\sqrt{2x^2+x-1}\right)+C\text{, }$
Which on substitution for $x$ gives (1). ]