Question

1+sinx÷1+cosx integrate​

Answer 1

GIVEN :–

$\\ \bf \implies \int \dfrac{1 + \sin(x)}{1 + \cos(x)} dx$

TO FIND :–

• Value of integration = ?

SOLUTION :–

• Let –

$\\ \implies\bf I = \int \dfrac{1 + \sin(x)}{1 + \cos(x)}.dx$

• Rationalization of denominator –

$\\ \implies\bf I = \int \dfrac{(1 + \sin x)(1 - \cos x)}{(1 + \cos x)(1 - \cos x)}.dx$

$\\ \implies\bf I = \int \dfrac{(1 + \sin x)(1 - \cos x)}{1 - \cos^{2}x}.dx$

$\\ \implies\bf I = \int \dfrac{1 +\sin x - \cos x - \sin x\cos x}{\sin^{2}x}.dx$

$\\ \implies\bf I = \int \bigg( \dfrac{1}{\sin^{2}x} + \dfrac{ \sin x}{\sin^{2}x} - \dfrac{ \cos x}{\sin^{2}x} - \dfrac{\sin x\cos x}{\sin^{2}x} \bigg).dx$

$\\ \implies\bf I = \int \bigg(cosec ^{2}x+cosec x- \cot x.cosec x - \cot x \bigg)dx$

$\\ \implies\bf I = \int cosec ^{2}x.dx+ \int cosec(x).dx- \int\cot x.cosec x .dx- \int \cot x.dx$

$\\ \large \implies{ \boxed{\bf I = - \cot(x) - \ln | cosec(x) + \cot(x) | +cosec(x) - \ln| \sin(x) |+ c}}$

• USED FORMULA :–

$\\ \longrightarrow \bf \int cosec ^{2}x.dx = - \cot(x) + c$

$\\ \longrightarrow \bf \int cosec(x).dx = - \ln | cosec(x) + \cot(x)|+ c$

$\\ \longrightarrow \bf \int \cot x.cosec x .dx = - cosec(x)+c$

$\\ \longrightarrow \bf \int \cot x = \ln| \sin(x) |+c$