Question

integrate 2cosx/(1-sinx)(2-cos²x)​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(2-cos^2x)}\,dx}$

$\underline{\textbf{To find:}}$

$\mathsf{\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(2-cos^2x)}\,dx}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(2-cos^2x)}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(1+(1-cos^2x))}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(1+sin^2x)}\,dx}$

$\boxed{\mathsf{Take,\;\;t=sinx\;\implies\;\dfrac{dt}{dx}=cosx\;\implies\;dt=cosx\,dx}}$

$\mathsf{=\displaystyle\int\,\dfrac{2}{(1-t)(1+t^2)}\,dt}$

$\mathsf{This\;can\;be\;written\;as}$

$\mathsf{=\displaystyle\int\,\left(\dfrac{t+1}{1+t^2}+\dfrac{1}{1-t}\right)\,dt}$

$\mathsf{=\displaystyle\int\,\left(\dfrac{t+1}{1+t^2}+\dfrac{1}{1-t}\right)\,dt}$

$\mathsf{=\displaystyle\int\,\dfrac{t}{1+t^2}dt+\int\,\dfrac{1}{1+t^2}dt+\int\,\dfrac{1}{1-t}\,dt}$

$\mathsf{=\displaystyle\dfrac{1}{2}\int\,\dfrac{2t}{1+t^2}dt+\int\,\dfrac{1}{1+t^2}dt-\int\,\dfrac{-1}{1-t}\,dt}$

$\mathsf{=\dfrac{1}{2}log(1+t^2)+tan^{-1}t+log(1-t)+C}$

$\mathsf{=\dfrac{1}{2}log(1+sin^2x)+tan^{-1}(sinx)+log(1-sinx)+C}$

$\underline{\textbf{Find more:}}$

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Answer 2

Answer:

2cosx

dx

\underline{\textbf{To find:}}

To find:

\mathsf{\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(2-cos^2x)}\,dx}∫

(1−sinx)(2−cos

2

x)

2cosx

dx

\underline{\textbf{Solution:}}

Solution:

\mathsf{Consider,}Consider,

\mathsf{\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(2-cos^2x)}\,dx}∫

(1−sinx)(2−cos

2

x)

2cosx

dx

\mathsf{=\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(1+(1-cos^2x))}\,dx}=∫

(1−sinx)(1+(1−cos

2

x))

2cosx

dx

\mathsf{=\displaystyle\int\,\dfrac{2\,cosx}{(1-sinx)(1+sin^2x)}\,dx}=∫

(1−sinx)(1+sin

2

x)

2cosx

dx

\boxed{\mathsf{Take,\;\;t=sinx\;\implies\;\dfrac{dt}{dx}=cosx\;\implies\;dt=cosx\,dx}}

Take,t=sinx⟹

dx

dt

=cosx⟹dt=cosxdx

\mathsf{=\displaystyle\int\,\dfrac{2}{(1-t)(1+t^2)}\,dt}=∫

(1−t)(1+t

2

)

2

dt

\mathsf{This\;can\;be\;written\;as}Thiscanbewrittenas

\mathsf{=\displaystyle\int\,\left(\dfrac{t+1}{1+t^2}+\dfrac{1}{1-t}\right)\,dt}=∫(

1+t

2

t+1

+

1−t

1

)dt

\mathsf{=\displaystyle\int\,\left(\dfrac{t+1}{1+t^2}+\dfrac{1}{1-t}\right)\,dt}=∫(

1+t

2

t+1

+

1−t

1

)dt

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