Question

Find the integration of the given function. I mark brainlist who gave me the answer plzzz solve it.. ​

Answer 1

Solution :

Now, $\bold{\int \frac{sin2x}{a^{2}sin^{2}x+b^{2}cos^{2}x}dx}$

= $\bold{\int \frac{2sinx\:cosx}{a^{2}sin^{2}x+b^{2}(1-sin^{2}x)}dx}$

    { since $\bold{sin^{2}x+cos^{2}x=1}$ and

    sin2x = 2 sinx cosx }

= $\bold{\int \frac{2sinx\:cosx}{b^{2}+(a^{2}-b^{2})sin^{2}x}dx}$

    { Let, sin²x = z

    Then, 2 sinx cosx dx = dz }

= $\bold{\int \frac{dz}{b^{2}+(a^{2}-b^{2})z^{2}}}$

= $\bold{\frac{1}{a^{2}-b^{2}} \int \frac{dz}{\frac{b^{2}}{a^{2}-b^{2}}+z^{2}}}$

= $\bold{\frac{1}{a^{2}-b^{2}} \int \frac{dz}{\bigg(\sqrt{\frac{b^{2}}{a^{2}-b^{2}}}\bigg)^{2}+z^{2}}}$

= $\bold{\frac{1}{a^{2}-b^{2}} \frac{1}{\sqrt{\frac{b^{2}}{a^{2}-b^{2}}}} \bigg(tan^{-1}\frac{z}{\sqrt{\frac{b^{2}}{a^{2}-b^{2}}}}\bigg) +C}$

where C is integral constant

= $\boxed{\small{\bold{\frac{1}{b\sqrt{a^{2}-b^{2}}} \bigg(tan^{-1} \frac{sin^{2}x}{\sqrt{\frac{b^{2}}{a^{2}-b^{2}}}}\bigg)+C}}}$ ,

    { since sin²x = z }

which is the required integral.

Answer 2

Answer:

$\frac{\log{|a^2\sin^2x+b^2\cos^2x|}}{a^2-b^2}+c$

Step-by-step explanation:

Let,

$I=\int{\frac{\sin2x}{a^2\sin^2s+b^2\cos^2x}}\,dx$

Let that,

$a^2\sin^2x+b^2\cos^2x=t\\\;\\\text{Difff. both sides with respect to x}\\\;\\a^2.2\sin x(cos x)+b^2.2\cos x.(-\sin x))=\frac{dt}{dx}\\\;\\a^2\sin2x-b^2\sin2x=\frac{dt}{dx}\\\;\\\sin2x(a^2-b^2)=\frac{dt}{dx}\\\;\\\sin2x.dx=\frac{dt}{a^2-b^2}$

Now,

$I=\frac{1}{a^2-b^2}\int{\frac{dt}{t}}\\\;\\I=\frac{1}{a^2-b^2}.\log{t}+c\\\;\\\text{Putting value of t}\\\;\\I=\frac{\log{|a^2\sin^2x+b^2\cos^2x|}}{a^2-b^2}+c$