Integration!

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

(A) x3x4dx\displaystyle \int { \sqrt { \frac { x-3 }{ x-4 } } dx }

Answer- (x3)(x4)+ln(x3+x4)+C\sqrt { (x-3)(x-4) } +\ln { \left( \sqrt { x-3 } +\sqrt { x-4 } \right) } +C

(B) dx[(x1)(2x)]32\displaystyle \int { \frac { dx }{ { \left[ \left( x-1 \right) \left( 2-x \right) \right] }^{ \frac { 3 }{ 2 } } } }

Answer- 2(x12x2xx1)+C2\left( \sqrt { \frac { x-1 }{ 2-x } } -\sqrt { \frac { 2-x }{ x-1 } } \right) +C

(C) dx[(x+2)8(x1)6]17dx\displaystyle \int { \frac { dx }{ { \left[ { \left( x+2 \right) }^{ 8 }{ \left( x-1 \right) }^{ 6 } \right] }^{ \frac { 1 }{ 7 } } } } dx

Answer- 73(x1x+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=sinmxcosnxdx{ I }_{ m,n }=\displaystyle \int { \sin ^{ m }{ x } \cos ^{ n }{ x } dx } then prove that Im,n=sinm+1xcosn1xm+n+n1m+n.Im,n2{ 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+6ex9ex4exdx=3536ln9e2x432x+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

Note by Anandhu Raj
3 years, 6 months ago

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

x3x4dx\displaystyle \int { \sqrt { \frac { x-3 }{ x-4 } } dx }

Take x4=u,dudx=12x4\sqrt{x-4}=u,\frac{du}{dx}=\frac{1}{2\sqrt{x-4}}

The integral becomes,

2u2+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 - 3 years, 6 months ago

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Yeah! Thank you! :)

Anandhu Raj - 3 years, 6 months ago

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Part B :\text{Part B :}

Put x1=t2    dx=2tdtx-1=t^2\implies dx=2t dt

I=2tdtt3(1t2)3/2    I=2dt1t21t2(1t2)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=sinθ    dθ=dt1t2    I=2dθsin2θ+cos2θ    I=2sin2θ+cos2θsin2θcos2θdθt=\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(sec2θ+csc2θ)dθ    I=2(tanθcotθ)+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 - 3 years, 6 months ago

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