Integration!

Here are some questions, can you please help me solve them?

(A) ∫x−3x−4dx\displaystyle \int { \sqrt { \frac { x-3 }{ x-4 } } dx }

Answer- (x−3)(x−4)+ln⁡(x−3+x−4)+C\sqrt { (x-3)(x-4) } +\ln { \left( \sqrt { x-3 } +\sqrt { x-4 } \right) } +C

(B) ∫dx[(x−1)(2−x)]32\displaystyle \int { \frac { dx }{ { \left[ \left( x-1 \right) \left( 2-x \right) \right] }^{ \frac { 3 }{ 2 } } } }

Answer- 2(x−12−x−2−xx−1)+C2\left( \sqrt { \frac { x-1 }{ 2-x } } -\sqrt { \frac { 2-x }{ x-1 } } \right) +C

(C) ∫dx[(x+2)8(x−1)6]17dx\displaystyle \int { \frac { dx }{ { \left[ { \left( x+2 \right) }^{ 8 }{ \left( x-1 \right) }^{ 6 } \right] }^{ \frac { 1 }{ 7 } } } } dx

Answer- 73(x−1x+2)1/7\frac { 7 }{ 3 } { \left( \frac { x-1 }{ x+2 } \right) }^{ 1/7 }

(D) Deduce the reduction formula for In=∫dx(1+x4)n{ I }_{ n }=\displaystyle \int { \frac { dx }{ { \left( 1+{ x }^{ 4 } \right) }^{ n } } } and hence evaluate I2{ I }_{ 2 }.

(E) If Im,n=∫sin⁡mxcos⁡nxdx{ I }_{ m,n }=\displaystyle \int { \sin ^{ m }{ x } \cos ^{ n }{ x } dx } then prove that Im,n=sin⁡m+1xcos⁡n−1xm+n+n−1m+n.Im,n−2{ I }_{ m,n }=\frac { \sin ^{ m+1 }{ x } \cos ^{ n-1 }{ x } }{ m+n } +\frac { n-1 }{ m+n } .{ I }_{ m,n-2 }

(F) Prove that ∫4ex+6e−x9ex−4e−xdx=3536ln⁡∣9e2x−4∣−32x+C\displaystyle \int { \frac { 4{ e }^{ x }+6{ e }^{ -x } }{ 9{ e }^{ x }-4{ e }^{ -x } } } dx=\frac { 35 }{ 36 } \ln { \left| 9{ e }^{ 2x }-4 \right| } -\frac { 3 }{ 2 } x+C

#Calculus

Note by Anandhu Raj
5 years, 2 months ago

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Comments

For part (A) here is how you can start-

∫x−3x−4dx\displaystyle \int { \sqrt { \frac { x-3 }{ x-4 } } dx }

Take x−4=u,dudx=12x−4\sqrt{x-4}=u,\frac{du}{dx}=\frac{1}{2\sqrt{x-4}}

The integral becomes,

2∫u2+1dx\displaystyle 2 \int { \sqrt {u^2+1 } dx }

Now substitute u=tan⁡(m)u=\tan (m),

I think from here you can do it on your own.

Akshay Yadav - 5 years, 2 months ago

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Yeah! Thank you! :)

Anandhu Raj - 5 years, 2 months ago

Part B :\text{Part B :}

Put x−1=t2  ⟹  dx=2tdtx-1=t^2\implies dx=2t dt

I=∫2tdtt3(1−t2)3/2  ⟹  I=2∫dt1−t21t2(1−t2)I = \int \frac{2t dt}{t^3(1-t^2)^{3/2}}\implies I =2\int \frac{dt}{\sqrt{1-t^2}}\frac{1}{t^2(1-t^2)}

Put t=sin⁡θ  ⟹  dθ=dt1−t2  ⟹  I=2∫dθsin⁡2θ+cos⁡2θ  ⟹  I=2∫sin⁡2θ+cos⁡2θsin⁡2θcos⁡2θdθt=\sin \theta\implies d\theta = \frac{dt}{\sqrt{1-t^2}}\implies I = 2\int \frac{d\theta}{\sin^2 \theta+\cos^2 \theta}\implies I = 2\int \frac{\sin^2 \theta+ \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}d\theta

I=2∫(sec⁡2θ+csc⁡2θ)dθ  ⟹  I=2(tan⁡θ−cot⁡θ)+CI = 2\int(\sec^2 \theta + \csc^2 \theta) d\theta \implies I = 2(\tan \theta - \cot \theta) + C

Substitute back θ,t\theta,t & get the desired result. Hope it helps

Aditya Narayan Sharma - 5 years, 2 months ago
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