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Category : JEE Main & Advanced

To evaluate such form of integrals, proceed as follows:

(1) Put \[\cos x=\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}\text{  and  }\sin x=\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}.\]

(2) Replace \[1+{{\tan }^{2}}\frac{x}{2}\]in the numerator by \[{{\sec }^{2}}\frac{x}{2}.\]

(3) Put \[\tan \frac{x}{2}=t\] so that \[\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt.\]

(4) Now, evaluate the integral obtained which will be of the form\[\int{\frac{dt}{a{{t}^{2}}+bt+c}}\] by the method discussed earlier.

(i) \[\int{\frac{dx}{a+b\cos x}}\]

Case I : When \[a>b\],  then

\[\int{\frac{dx}{a+b\cos x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\,\,{{\tan }^{-1}}\left( \sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} \right)}+c\]

Case II : When \[a<b\], then

\[\int{\frac{dx}{a+b\cos x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b-a}\tan \frac{x}{2}+\sqrt{b+a}}{\sqrt{b-a}\tan \frac{x}{2}-\sqrt{b+a}} \right|+c\]

Case III : When \[a=b\], then \[\int{\frac{dx}{a+b\cos x}}=\frac{1}{a}\tan \frac{x}{2}+c\].

(ii) \[\int{\frac{dx}{a+b\sin x}}\]

Case I : When \[{{a}^{2}}>{{b}^{2}}\] or \[a>0\] and \[a>b\],  then

\[\int{\frac{dx}{a+b\sin x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}}{{\tan }^{-1}}\left[ \frac{a\tan \frac{x}{2}+b}{\sqrt{{{a}^{2}}-{{b}^{2}}}} \right]+c\]

Case II : When \[{{a}^{2}}<{{b}^{2}}\], then

\[\int{\frac{dx}{a+b\sin x}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}}\log \left| \frac{(a\tan \frac{x}{2}+b)-(\sqrt{{{b}^{2}}-{{a}^{2}}})}{(a\tan \frac{x}{2}+b)+\sqrt{{{b}^{2}}-{{a}^{2}}}} \right|+c\]

Case III : When \[{{a}^{2}}={{b}^{2}}\]

In this case, either \[b=a\] or \[b=-a\]

(a) When \[b=a\], then 

\[\int{\frac{dx}{a+b\sin x}}=\frac{-1}{a}\cot \left( \frac{\pi }{4}+\frac{x}{2} \right)\,+c=\frac{1}{a}[\tan x-\sec x]+c\]

(b) When \[b=-a\], then \[\int{\frac{dx}{a+b\sin x}=\frac{1}{a}\tan \left( \frac{\pi }{4}+\frac{x}{2} \right)}+c\].