Question

$\\ \displaystyle\Huge\int\sf\tiny\dfrac{\Big\{cos^{-1}x\big[\sqrt{(1-x^2)}\:\big]\Big\}^{-1}}{log_{e}\left\{1+\left[\dfrac{sin\left\lgroup 2x\sqrt{(1-x^2)}\right\rgroup}{\boldsymbol\pi}\right]\right\}}dx$

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Answer 1

Answer:

Answer :

$\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}$

Step by Step Explanation :

$\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}$

➳ Put x = sinθ

∴ dx = cosθ dθ

$\therefore\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-x^2}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-\sin ^2\theta}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{\cos ^2\left(\theta\right)}}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \cos \left(\theta\right) }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\cos \:^{-1}\left(\cos \:\left(\dfrac{\pi }{2}+\theta \:\right)\right)\:\cdot \:\cos \theta }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\:\left(\dfrac{\pi \:}{2}+\theta \:\right)\:\cdot \:\cos \theta }$

Also

$\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=\sin ^{-1}(2 \sin \theta \cos \theta)}$

$\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=sin^{-1}\left(sin2\theta \right)}$

$\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=2\theta}$

Now

$\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}$

$\sf{=\displaystyle\int \frac{\left(\cos ^{-1} \sin \theta \sqrt{1-\sin ^{2} \theta}\right)^{-1}}{log_e \left(1+\frac{2 \theta}{\pi}\right)} \cos \theta d \theta}$

$\sf{\displaystyle=\int\frac{\left(\frac{\pi }{2}+\theta\right)^{-1}\:\cdot \:\cos \:^{-1}\:\theta \:\cdot \:\cos \:\theta \:d\:\theta }{\log _e\left(\frac{\pi +2\:\theta }{\pi }\right)}}$

$\sf{\displaystyle=\int \:\frac{2d\theta }{\left(\pi +2\theta \right)\:\log_e\left(\frac{\pi +2\theta }{\pi }\right)}}$

➳ Put  $\sf{log\left(\dfrac{\pi +2\theta }{\pi }\right)=t}$

$\to\sf{\dfrac{\pi}{\pi+2 \theta} d \theta=d t}$

$\to\sf{\dfrac{2}{\pi} \int \dfrac{d t}{t}}$

$\to\sf{\dfrac{2}{\pi} \log (t)}$

➳Substitute back the Value of t

$\to\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}$

$\boxed{\boxed{\therefore\:\:\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}=\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}}}$

Answer 2

Answer:

2

log

e

(log

e

(

π

π+2θ

))

Step by Step Explanation :

\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}∫

log

e

{1+[

π

sin(2x

(1−x

2

)

)

])

{cos

−1

x[

(1−x

2

)

]}

−1

dx

➳ Put x = sinθ

∴ dx = cosθ dθ

\therefore\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-x^2}\right\} }∴cos

−1

x⋅{

1−x

2

}=cos

−1

(sinθ)⋅{

1−x

2

}

\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-\sin ^2\theta}\right\} }→cos

−1

x⋅{

1−x

2

}=cos

−1

(sinθ)⋅{

1−sin

2

θ

}

\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{\cos ^2\left(\theta\right)}}\right\} }

\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \cos \left(\theta\right) }→cos

−1

x⋅{

1−x

2

}=cos

−1

(sinθ)⋅cos(θ)

\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\cos \:^{-1}\left(\cos \:\left(\dfrac{\pi }{2}+\theta \:\right)\right)\:\cdot \:\cos \theta }→cos

−1

x⋅{

1−x

2

}=cos

−1

(cos(

2

π

+θ))⋅cosθ

\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\:\left(\dfrac{\pi \:}{2}+\theta \:\right)\:\cdot \:\cos \theta }→cos

−1

x⋅{

1−x

2

}=(

2

π

+θ)⋅cosθ

Also

\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=\sin ^{-1}(2 \sin \theta \cos \theta)}sin

−1

(2x

1−2x

2

)=sin

−1

(2sinθcosθ)

\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=sin^{-1}\left(sin2\theta \right)}→sin

−1

(2x

1−2x

2

)=sin

−1

(sin2θ)

\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=2\theta}→sin

−1

(2x

1−2x

2

)=2θ

Now

\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}∫

log

e

{1+[

π

sin(2x

(1−x

2

)

)

])

{cos

−1

x[

(1−x

2

)

]}

−1

dx

\sf{=\displaystyle\int \frac{\left(\cos ^{-1} \sin \theta \sqrt{1-\sin ^{2} \theta}\right)^{-1}}{log_e \left(1+\frac{2 \theta}{\pi}\right)} \cos \theta d \theta}=∫

log

e

(1+

π

2θ

)

(cos

−1

sinθ

1−sin

2

θ

)

−1

cosθdθ

\sf{\displaystyle=\int\frac{\left(\frac{\pi }{2}+\theta\right)^{-1}\:\cdot \:\cos \:^{-1}\:\theta \:\cdot \:\cos \:\theta \:d\:\theta }{\log _e\left(\frac{\pi +2\:\theta }{\pi }\right)}}=∫

log

e

(

π

π+2θ

)

(

2

π

+θ)

−1

⋅cos

−1

θ⋅cosθdθ

\sf{\displaystyle=\int \:\frac{2d\theta }{\left(\pi +2\theta \right)\:\log_e\left(\frac{\pi +2\theta }{\pi }\right)}}=∫

(π+2θ)log

e

(

π

π+2θ

)

2dθ

➳ Put \sf{log\left(\dfrac{\pi +2\theta }{\pi }\right)=t}log(

π

π+2θ

)=t

\to\sf{\dfrac{\pi}{\pi+2 \theta} d \theta=d t}→

π+2θ

π

dθ=dt

\to\sf{\dfrac{2}{\pi} \int \dfrac{d t}{t}}→

π

2

∫

t

dt

\to\sf{\dfrac{2}{\pi} \log (t)}→

π

2

log(t)

➳Substitute back the Value of t

\to\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}→

π

2

log

e

(log

e

(

π

π+2θ

))

\boxed{\boxed{\therefore\:\:\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}=\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}}}

∴∫

log

e

{1+[

π

sin(2x

(1−x

2

)

)

])

{cos

−1

x[

(1−x

2

)

]}

−1

dx=

π

2

log

e

(log

e

(

π

π+2θ

))