find the integral of :
(sec x . dx)/{ sin(2x+A) + sin A}1/2
a) secA. {tanx.sinA +cosA)1/2 + K (K-constant here)
b)21/2 .sec A {tanA +sinAcosA)1/2 + K
c) 2-1/2.secA (tanA.sinA + cos A)3/2 + K
d) none of these
\(\int {{{Secx}\over {\sqrt {Sin(2x+A)+SinA}}}dx}\)
\(=\int {{{Secx}\over {\sqrt {2Sin(x+A)Cosx}}}dx}\)
\(=\int {{{Sec^2 x}\over {\sqrt {2TanxCosA+2SinA}}}dx}\)
\(={1\over {\sqrt{2}CosA}}\int {{{Sec^2xCosA}\over {\sqrt {TanxCosA+SinA} }}dx}\)
\(={1 \over {\sqrt{2}CosA}}\int {{{f'(x)}\over {\sqrt {f(x)} }}dx}\)
\(={1 \over {\sqrt{2}CosA}}2\sqrt {f(x)}+C\)
\(=\sqrt{2}SecA\sqrt {TanxCosA + SinA}+C\)
\(=\int {{{Secx}\over {\sqrt {2Sin(x+A)Cosx}}}dx}\)
\(=\int {{{Sec^2 x}\over {\sqrt {2TanxCosA+2SinA}}}dx}\)
\(={1\over {\sqrt{2}CosA}}\int {{{Sec^2xCosA}\over {\sqrt {TanxCosA+SinA} }}dx}\)
\(={1 \over {\sqrt{2}CosA}}\int {{{f'(x)}\over {\sqrt {f(x)} }}dx}\)
\(={1 \over {\sqrt{2}CosA}}2\sqrt {f(x)}+C\)
\(=\sqrt{2}SecA\sqrt {TanxCosA + SinA}+C\)
An easy way to solve the given question is:
See whether the function given is even or odd. If the function is even , its integral will be odd and vice versa.
HINTS FOR THE GIVEN INTEGRAL
1.Here in the question for simplification , we can put value of A=0.
2.Then the function reduces to
(sec x . dx)/{ sin2x }1/2
Putting A=0 in the given choices we will get
a) 1 + K
b)0 + K
c) 2-1/2 + K
d) none of these
Hence the correct choice is d) because the integral should be in the form of x.