2
$\begingroup$

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

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)

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

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 $\arcsin$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.