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I am currently working through trigonometric integrals and working on this problem specifically:

$$ \int_{}^{} \sin ^{2}x\cdot\sin (2x) \, dx $$

And the way I approached this was to convert $\sin ^{2}x$ to $\frac{1}{2}(1-cos(2x)$ using that trigonometric identity and working through the problem as follows:

$$ \int_{}^{} \sin ^{2}x\cdot\sin (2x) \, dx \\ \\ $$ $$ =\int_{}^{} \frac{1}{2}\cdot(1-\cos(2x))\cdot \sin (2x) \, dx \\ \\ $$ $$ =\frac{1}{2}\int_{}^{} (1-\cos(2x))\cdot \sin (2x) \, dx \\ \\ $$ $$ =\frac{1}{2}\int_{}^{} \sin(2x)-\sin(2x)\cdot \cos(2x) \, dx \\ \\ $$ $$ =\frac{1}{2}\int_{}^{} \sin(2x) \, dx - \frac{1}{2}\int_{}^{} \sin(2x)\cdot \cos(2x) \, dx \\ $$ $$ \text{let u = sin(2x)} \\ $$ $$ \text{therefore du = 2cos(2x) dx} \\ \\ $$ $$ =\frac{1}{2}\int_{}^{} \sin(2x) \, dx -\frac{1}{2}\int_{}^{} \frac{1}{2}u \, du \\ \\ $$ $$ =-\frac{1}{4}\cos(2x) -\frac{1}{8}u^{2} \\ \\ $$ $$ =-\frac{1}{4}\cos(2x) - \frac{1}{8}\sin ^{2} (2x) + C $$

But the textbook's answer is:

$$\frac{1}{2}sin ^{4}(x) + C$$

and I guess the way they've approached the problem is by using the trigonometric identity $sin(2x) = 2sin(x)cos(x)$ and then u-subbing out sin(x) from the resulting:

$$ \int \sin ^{2}x(2\sin x\cos x) dx $$ and working through that (which admittedly is way more elegant/faster). But what I'm wondering is why my answer seems correct to me but is so different and even graphs differently than the textbook's answer?

Conjonorama
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1 Answers1

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Your answer is also correct since $$ -\frac14\cos(2x)-\frac18\sin^2(2x)+\frac14 = \frac12\sin^4(x). $$ That is, \begin{align} -\frac14\cos(2x)-\frac18\sin^2(2x)+C_1 &= -\frac14\cos(2x)-\frac18\sin^2(2x)+\frac14+C_2 \\ &= \frac12\sin^4(x)+C_2 \end{align} where $C_2 := C_1-1/4$.

azif00
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  • Thank you! I feel like that makes sense intuitively. I guess I have trouble grasping where if this were a definite integral would I get the same 'output' if I had approached the problem separate ways? Like with my solution and the textbook's solution over the interval [0, Pi/2] both solutions are positive 1/2. But when we add the constant of 1/4 that changes my answer now by 1/4. I think this is mostly my lack of a deeper understanding to be honest. – Conjonorama May 14 '25 at 03:21
  • @Conjonorama When you compute a definite integral, $\int_a^b f(x)dx$, you need to find some antiderivative of $f(x)$, say $F(x)$, and then $\int_a^b f(x)dx = F(b)-F(a)$. Note that, if $c$ is a constant, $G(x)=F(x)+c$ is also an antiderivative of $f(x)$, but $$ G(b)-G(a) = F(b)+c-[F(a)+c] = F(b)-F(a). $$ In other words, no matter which antiderivative you choose, that difference is always the same number. – azif00 May 14 '25 at 03:30