What substitution should I make?

Calculus Level 3

∫ 0 1 x 6 ( 1 − x 2 ) 1 2 d x \large \int_0^1 x^6 (1-x^2)^{\frac12} \, dx

Given that the integral above is equal to A B π \dfrac AB \pi for coprime positive integers A A and B B , find the value of A + B A+B .


The answer is 261.

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Mahimn Bhatt by Mahimn Bhatt , 23, India

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1 solution

Let x = sin ⁡ θ x = \sin {\theta} , then we have:

∫ 0 1 x 6 1 − x 2 d x = ∫ 0 π 2 sin ⁡ 6 θ 1 − sin ⁡ 2 θ cos ⁡ θ d θ = ∫ 0 π 2 sin ⁡ 6 θ cos ⁡ 2 θ d θ \displaystyle \int _0 ^1 {x^6\sqrt{1-x^2} dx} = \int _0 ^{\frac{\pi}{2}} {\sin^6 {\theta} \sqrt{1-\sin^2{\theta}} \cos{\theta} d\theta} = \int _0 ^{\frac{\pi}{2}} {\sin^6 {\theta} \cos^2 {\theta} d\theta}

= 1 4 ∫ 0 π 2 sin ⁡ 4 θ sin ⁡ 2 2 θ d θ = 1 16 ∫ 0 π 2 ( 1 − cos ⁡ 2 θ ) 2 ( 1 − cos ⁡ 2 2 θ ) d θ \displaystyle = \frac {1}{4} \int _0 ^{\frac{\pi}{2}} {\sin^4 {\theta} \sin^2 {2\theta} d\theta} = \frac {1}{16} \int _0 ^{\frac{\pi}{2}} {(1-\cos{2\theta})^2 (1- \cos^2 {2\theta}) d\theta}

= 1 16 ∫ 0 π 2 ( 1 − cos ⁡ 2 θ ) 3 ( 1 + cos ⁡ 2 θ ) d θ \displaystyle = \frac {1}{16} \int _0 ^{\frac{\pi}{2}} {(1-\cos{2\theta})^3 (1+ \cos {2\theta}) d\theta}

= 1 16 ∫ 0 π 2 ( 1 − 3 cos ⁡ 2 θ + 3 cos ⁡ 2 2 θ − cos ⁡ 3 2 θ ) ( 1 + cos ⁡ 2 θ ) d θ \displaystyle = \frac {1}{16} \int _0 ^{\frac{\pi}{2}} {(1-3\cos{2\theta} + 3\cos^2{2\theta} - \cos^3{2\theta}) (1+ \cos {2\theta}) d\theta}

= 1 16 ∫ 0 π 2 ( 1 − 2 cos ⁡ 2 θ + 2 cos ⁡ 3 2 θ − cos ⁡ 4 2 θ ) d θ \displaystyle = \frac {1}{16} \int _0 ^{\frac{\pi}{2}} {(1-2\cos{2\theta} + 2\cos^3{2\theta} - \cos^4{2\theta}) d\theta}

= 1 16 [ θ − sin ⁡ 2 θ + sin ⁡ 2 θ − sin ⁡ 3 2 θ 3 ] 0 π 2 − 1 16 ∫ 0 π 2 cos ⁡ 4 2 θ d θ \displaystyle = \frac {1}{16} \left[ \theta - \sin{2\theta} + \sin{2\theta} - \frac {\sin^3{2\theta}}{3} \right] _0^{\frac{\pi}{2}} - \frac {1}{16} \int _0 ^{\frac{\pi}{2}} {\cos^4{2\theta} d\theta}

= π 32 − 1 64 ∫ 0 π 2 ( 1 + cos ⁡ 4 θ ) 2 d θ \displaystyle = \frac {\pi}{32} - \frac {1}{64} \int _0 ^{\frac{\pi}{2}} {(1+\cos{4\theta})^2 d\theta}

= π 32 − 1 64 ∫ 0 π 2 ( 1 + 2 cos ⁡ 4 θ + cos ⁡ 2 4 θ ) d θ \displaystyle = \frac {\pi}{32} - \frac {1}{64} \int _0 ^{\frac{\pi}{2}} {(1+2 \cos{4\theta} + \cos^2{4\theta}) d\theta}

= π 32 − 1 64 [ θ + sin ⁡ 4 θ 2 + θ 2 + sin ⁡ 8 θ 16 ] 0 π 2 \displaystyle = \frac {\pi}{32} - \frac {1}{64} \left[\theta + \frac {\sin{4\theta}}{2} + \frac {\theta}{2} + \frac {\sin{8\theta}}{16} \right] _0 ^{\frac{\pi}{2}}

= π 32 − 1 64 [ 3 π 4 ] = 5 π 256 \displaystyle = \frac {\pi}{32} - \frac {1}{64} \left[ \frac{3\pi}{4} \right] = \boxed{\dfrac{5\pi}{256}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Good solution sir.!!!! Sir there is another method through which i have solved. Using beta function. Putting x^2 = t and generating beta integral. Kindly comment. on my method..!

Mahimn Bhatt - 6 years, 4 months ago

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Oh, I was sure there must be an easier way to solve this problem. Thanks for the hint. Learnt beta function 30 years ago but never got to use it. The solution is here:

∫ 0 π 2 sin ⁡ n θ cos ⁡ m θ d θ − 1 2 B ( 1 2 ( n + 1 ) , 1 2 ( m + 1 ) ) \displaystyle \int _0 ^{\frac{\pi}{2}} {\sin^n {\theta} \cos ^m {\theta} d\theta} - \frac {1}{2} B \left( \frac {1}{2} (n+1), \frac {1}{2} (m+1) \right)

⇒ ∫ 0 π 2 sin ⁡ 6 θ cos ⁡ 2 θ d θ − 1 2 B ( 7 2 , 3 2 ) = Γ ( 7 2 ) Γ ( 3 2 ) 2 Γ ( 7 2 + 3 2 ) \displaystyle \Rightarrow \int _0 ^{\frac{\pi}{2}} {\sin^6 {\theta} \cos ^2 {\theta} d\theta} - \frac {1}{2} B \left( \frac {7}{2}, \frac {3}{2} \right) = \frac {\Gamma (\frac {7}{2}) \Gamma (\frac {3}{2})}{2\Gamma (\frac {7}{2}+\frac {3}{2})}

= 5 2 ˙ 3 2 ˙ 1 2 ˙ π ˙ 1 2 ˙ π 2 ˙ 4 ! = 5 π 256 \displaystyle \quad \space = \frac {\frac{5}{2}\dot{} \frac{3}{2}\dot{} \frac{1}{2}\dot{} \sqrt{\pi} \dot{} \frac{1}{2}\dot{} \sqrt{\pi}}{2 \dot{} 4!} = \boxed{\dfrac {5\pi}{256}}

Chew-Seong Cheong - 6 years, 4 months ago

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Thanks sir.!

Mahimn Bhatt - 6 years, 4 months ago

Another method: Apply the reduction formula:

∫ 0 π / 2 sin ⁡ 2 n θ d θ = ( 2 n − 1 ) ! ! ( 2 n ) ! ! × π 2 . \int_0^{\pi/2} \sin^{2n} \theta \, d\theta = \dfrac{(2n-1)!!}{(2n)!!} \times \dfrac \pi 2.

So on your very first line, we have

… = ∫ 0 π / 2 sin ⁡ 6 θ ( 1 − sin ⁡ 2 θ ) d θ = ∫ 0 π / 2 sin ⁡ 6 θ d θ − ∫ 0 π / 2 sin ⁡ 8 θ d θ = π 2 ( 5 ! ! 6 ! ! − 7 ! ! 8 ! ! ) = 5 π 256 \ldots = \int_0^{\pi/2} \sin^6\theta (1 - \sin^2\theta) \, d\theta = \int_0^{\pi/2} \sin^6\theta \, d\theta- \int_0^{\pi/2} \sin^8\theta \, d\theta= \dfrac\pi2 \left( \dfrac{5!!}{6!!} - \dfrac {7!!}{8!!} \right) = \dfrac{5\pi}{256}

Pi Han Goh - 5 years, 7 months ago

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