JEE Main & Advanced Mathematics Indefinite Integrals Integrals of the Form \[\int{\frac{dx}{a+bcosx+csinx}}\,,\,\,\int{\frac{dx}{asinx+bcosx}}\]

(1) Integral of the form \[\int{\frac{dx}{a+b\,cos\,x+csinx}}\]: To evaluate such integrals, we put \[b=r\cos \alpha \] and \[c=r\sin \alpha .\]

So that, \[{{r}^{2}}={{b}^{2}}+{{c}^{2}}\] and \[\alpha ={{\tan }^{-1}}\frac{c}{b}.\] 

\[\therefore \,\,\,I=\int{\frac{dx}{a+r(\cos \alpha \,\cos x+\sin \alpha \sin x)}}\]\[=\int{\frac{dx}{a+r\cos (x-\alpha )}}\]

Again, put \[x-\alpha =t\,\]\[\Rightarrow \,\,dx=dt,\] we have \[I=\int{\frac{dt}{a+r\cos t}}\]

Which can be evaluated by the method discussed earlier.

(2) Integral of the form \[\int{\frac{dx}{a\,sin\,x+b\,cos\,x}}\] : To evaluate this type of integrals we substitute \[a=r\cos \theta ,\] \[b=r\sin \theta \] and so \[r=\sqrt{{{a}^{2}}+{{b}^{2}}},\,\,\alpha ={{\tan }^{-1}}\frac{b}{a}\] .

So, \[\int{\frac{dx}{a\sin x+b\cos x}}=\frac{1}{r}\int{\frac{dx}{\sin (x+\alpha )}}=\frac{1}{r}\int{\text{cosec}(x+\alpha )dx}\]

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

The integral of the above form can be evaluated by using \[\cos x=\frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2}\] and \[\sin x=\frac{2\tan (x/2)}{1+{{\tan }^{2}}(x/2)}\].