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
$\int \sqrt{\frac{1+x}{2 x-1}} d x=\int \frac{1+x}{\sqrt{2 x^2+x-1}} d x=\frac{1}{4} \int \frac{4 x+1}{\sqrt{2 x^2+x-1}} d x+\frac{3}{4 \sqrt{2}} \int \frac{d x}{\sqrt{\left(x+\frac{1}{4}\right)^2-\frac{9}{16}}}$
$=\frac{1}{2} \sqrt{2 x^2+x-1}+\frac{3}{4 \sqrt{2}} \log \left(x+\frac{1}{4}+\sqrt{\left(x+\frac{1}{4}\right)^2-\frac{9}{16}}\right)+C$
$=\frac{1}{2} \sqrt{2 x^2+x-1}+\frac{3}{4 \sqrt{2}} \log \left(4 x+1+2 \sqrt{2} \sqrt{2 x^2+x-1}\right)+C \text {, }$
Which on substitution for $x$ gives (1). ]