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$$\int \sin^3(3x)\cos^{-2}(3x)dx$$

Let$u=3x$; then $du=3dx$, so $dx=\dfrac{du}{3}$

$$\dfrac{1}{3}\int \sin^3(u)\cos^{-2}(u)du$$

Expand $\sin^3(u)$ to $\sin^2(u)\sin(u)$

$$\dfrac{1}{3}\int \sin^2(u)\sin(u)\cos^{-2}(u)du$$

Re-write $\sin^2(u)$ to $(1-\cos^2(u))$

$$\dfrac{1}{3}\int (1-\cos^2(u))\cos^{-2}(u)\sin(u)du$$

Multiply $(1-\cos^2(u))$ by $\cos^{-2}(u)$

$$\dfrac{1}{3}\int (\cos^{-2}(u)-1)\sin(u)du$$

Let $v=\cos(u)$; then $dv=-\sin(u)du$, so $-dv=\sin(u)du$

$$-\dfrac{1}{3}\int (v^{-2}-1)dv$$

Integrate

$$-\dfrac{1}{3}(-v^{-1}-v)$$

Re-write in terms of $x$

$$-\dfrac{1}{3}(-\cos^{-1}(3x)-\cos(3x))+C$$

Incorrect answer according to MyMathLab (Pearson Education)

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Thanks to whoever commented; not sure why your comment disappeared. The problem was most likely with the $cos^{-1}$ notation. –  Evan Sep 19 '12 at 2:36
    
You might have skipped some steps by noticing $(\cos^{-2}u-1)\sin u=\sec u\tan u-\sin u$ –  Mike Sep 19 '12 at 4:02

1 Answer 1

up vote 2 down vote accepted

Your answer is correct. However, note that

$$-\dfrac{1}{3}(-\frac{1}{\cos(3x)}-\cos(3x)) = \dfrac{1}{3} (\cos(3x)+\sec(3x))$$

Differentiating your answer, we have:

$$\dfrac{1}{3} \frac{\text{d}}{\text{d}x} (\cos(3x)+\sec(3x))+C = \dfrac{1}{3} \cdot 3 (-\sin (3x)+\tan (3x) \sec (3x)) = \sin(3x)\tan^2(3x) = \frac{\sin^3 (3x)}{\cos^2 (3x)}$$

So your integral is correct.


Try and avoid the $\sin^{-1}$ notation for $\frac{1}{\sin x}$, as people may confuse with $\arcsin$.

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