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I have solved this problem by first converting $\cos x$ to $\sin x$ using identity $\sin^2 x + cos^2 x = 1$ and then used power reduction formula to integrate.

My answer is $-\frac{1}{48} \left(8\sin^5x -2\sin^3 x - 33 \sin x + \frac{11}{16} x \right)$

But the correct answer is $-\frac{1}{192} \left(\sin 6x -3 \sin 4x + \sin 2x + 12x \right)$

Why am I getting wrong answer? I think I am using right method. Please help.

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    Disclose ur steps – 2017-02-12
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    Use $\cos2x=1-2\sin^2x=2\cos^2x-1$ – 2017-02-12
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    hint :$$\int (sinx)^4 (cosx)^2 dx=\\ \int (\dfrac{1-cos2x}{2})^2 (\dfrac{1+cos2x}{2}) dx=\\ \dfrac18\int (1-cos2x)^2(1+cos2x)dx\\ \dfrac18\int (1-2cos2x+cos^22x)(1+cos2x)dx=\\ \dfrac18\int (1-2cos2x+\dfrac12(1+cos4x))(1+cos2x)dx=$$ – 2017-02-12

2 Answers 2

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Your approach is correct, and I see no reason to believe your answer isn't; while I haven't worked it out that way myself, it's clear that your approach would give an answer in that format.

The thing to be aware of is that trigonometric functions have complicated interrelations. For example, $\sin(2x) = 2\sin(x)\cos(x)$. So just because your answer looks different doesn't mean it is different. The fact that the integral has a $+c$ at the end, which can absorb constants, makes it very difficult to spot these equivalences.

I do think the $\frac{11}{16}$ in your answer is probably wrong - that's the coefficient of the $x$ term, so it won't be affected by any trig identities, but there's no factor of $11$ in the coefficient of the $x$ term in the correct answer. But that's probably just an algebra error you made in the power reduction formula. Double-check your steps.

Apart from that, though, I'd say you're fine. Certainly your approach is correct, and if executed correctly will give you a correct answer.

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$$\int(\sin^4 x)(\cos^2 x) dx$$

$$\cos^2x=1-\sin^2x$$

$$\int(\sin^4 x)(1-\sin^2x) dx=\int(\sin^4 x-\sin^6x)dx=\int\sin^4 x dx-\int\sin^6xdx$$

Integration by reduction formulae:

$$\bbox[yellow,5px] {I_n=\int\sin^nxdx => nI_n = -\sin^{n-1}{x}\cos{x}+(n-1)I_{n-2}\,\!}$$

$$\int\sin^6 x dx={-\sin^5x\cos x+5\int \sin^4xdx\over6}$$

$$\int\sin^4 x dx-\int\sin^6xdx=\int \sin^4xdx-{1\over6}({-\sin^5x\cos x+5 \int \sin^4xdx})={1\over6}({\sin^5x\cos x)-{5\over6} \int \sin^4xdx+\int \sin^4xdx}={1\over6}({\sin^5x\cos x)+{1\over6}\int \sin^4xdx}$$

$$\int\sin^4 x dx={-\sin^3x\cos x+3\int \sin^2xdx\over4}$$

$${1\over6}({\sin^5x\cos x)+{1\over6}\int \sin^4xdx}={1\over6}({\sin^5x\cos x)+{}{-\sin^3x\cos x+3\int \sin^2xdx\over24}}={1\over6}({\sin^5x\cos x)+{}{-\sin^3x\cos x+{3\over4} (2 x - \sin2 x)\over24}}={1\over6}({\sin^5x\cos x)+{-1\over24}{\sin^3x\cos x+{3\over96} (2 x - \sin2 x)}}$$$$=[{1\over192} (5 \sin2 x - 4 \sin4 x + \sin6 x)]+[{1\over192} (\sin4 x - 2 \sin2 x)]+[{3\over96} (2 x - \sin2 x)]={1\over192} (12 x - 3 \sin2 x - 3 \sin4 x + \sin6 x)+C$$