$$\int\sin^6(x)\cos(x)\,dx.$$
Apparently...
$$\int\sin^6(x)\cos(x)\,dx= \int\sin^6(x)\,d\sin(x)x = \frac{1}{7}\sin^7(x)+c.$$
Can somebody explain the second equality? Shouldn't there be a $dx$ after it?
$\endgroup$ 35 Answers
$\begingroup$Directly, using
$$\int f'(x) f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1}+C$$
with $\;f(x)=\sin x\;$ , in your case...
$\endgroup$ $\begingroup$The integral is a good example of substitution. $$ \int \sin^6(x)\cos(x)\;dx $$ Let $\color{green}{u=\sin(x)}$, then $\frac{du}{dx} = \cos(x)$, and we write this as $\color{blue}{du = \cos(x)dx}$. So $$ \int \color{green}{\sin(x)}^6\color{blue}{\cos(x)\;dx} = \int \color{green}{u}^6 \color{blue}{du} = \frac{1}{7}u^7 + C = \frac{1}{7}\sin^7(x) + C $$ In your notation you have $$ \int \sin^6(x) d\sin(x) $$ and the $d\sin(x)$ is sometimes used to indicate the change of variable that we have used substituting $u$ for $\sin(x)$. So the $du$ is like the $d\sin(x)$.
$\endgroup$ $\begingroup$Hint: That's what you have
$$ \int (f(x))^n f'(x)dx = \frac{(f(x))^{n+1}}{n+1}+c .$$
You can see this by using the substitution $u=\sin(x)$.
$\endgroup$ 2 $\begingroup$It looks like your second equality has $\sin x$ sandwiched between $d$ and $x$ as in "dx".
Correctly stated, we have:
$$\int\sin^6(x)\cos(x)\,dx= \int\sin^6(x)\,\dfrac{d}{dx}(\sin(x))\,dx = \frac{1}{7}\sin^7(x)+c.$$
$\endgroup$ 0 $\begingroup$$$ \frac{d}{dx} \sin x = \cos x $$ Hence $$ d\sin x = \cos x\,dx. $$
In other words, the infinitely small increment of $\sin x$ is equal to $\cos x$ times the infinitely small increment of $x$.
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