A Normal Distribution?

Calculus Level 4

Given that the indefinite integral

sin 5 2 x cos 3 x d x \large \int \sin^{\frac {5}{2}} x \cos^3 x \space \mathrm{d}x

equals to

2 sin A 2 x [ 1 B 1 C sin 2 x ] \large 2 \sin^{ \frac {A}{2} } x \left [ \frac {1}{B} - \frac {1}{C} \sin^2 x \right ]

neglecting the arbitrary constant.

Find the value of A + B C A + B - C .


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The answer is 3.

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2 solutions

Similar to Harshvardhan Mehta 's solution, but in LaTex here.

sin 5 2 x cos 3 x d x = sin 5 2 x cos 2 x d sin x = sin 5 2 x ( 1 sin 2 x ) d sin x = ( sin 5 2 x sin 9 2 x ) d sin x = 2 7 sin 7 2 x 2 11 sin 11 2 x = 2 sin 7 2 x [ 1 7 1 11 sin 2 x ] \displaystyle \begin{aligned} \int {\sin^{\frac{5}{2}}{x}\cos^3{x} \space dx} & = \int {\sin^{\frac{5}{2}}{x}\cos^2{x} \space d\sin{x}} \\ & = \int {\sin^{\frac{5}{2}}{x}(1- \sin^2{x}) \space d\sin{x}} \\ & = \int {(\sin^{\frac{5}{2}}{x}- \sin^{\frac{9}{2}}{x}) \space d\sin{x}} \\ & = \frac {2}{7} \sin^{\frac{7}{2}}{x}-\frac {2}{11} \sin^{\frac{11}{2}}{x} \\ & = 2 \sin^{\frac{7}{2}}{x} \left[ \frac{1}{7}-\frac{1}{11}\sin^2{x}\right] \end{aligned}

A + B C = 7 + 7 11 = 3 \Rightarrow A+B-C = 7+7-11 = \boxed{3}

thanks a lot sir for posting it in Latex .

Harshvardhan Mehta - 6 years, 3 months ago

nice method...

saket khandal - 5 years, 4 months ago

How come level 5?

U Z - 6 years, 3 months ago

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please reduce it to level 3.

Harshvardhan Mehta - 6 years, 3 months ago

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Sorry I can't do

U Z - 6 years, 3 months ago

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