Monster antiderivative for fun: $\displaystyle\int\dfrac{3e^{2x}\sin(x)+4e^{2x}\cos(x)}{-(\sin^2(x)+\cos^2(x)+\tan^2(x))}dx$

Question

So I was bored, and decided to do antiderivatives for fun. After a while, I came up with this:$$\int\dfrac{3e^{2x}\sin(x)+4e^{2x}\cos(x)}{-(\sin^2(x)+\cos^2(x)+\tan^2(x))}dx$$which I thought that I might be able to do. Here is my attempt at doing so:

So first we notice that we can rewrite the the bottom of the fraction as$$\sin^2(x)+\cos^2(x)+\tan^2(x)=1+\tan^2(x)=\sec^2(x)$$and then noting that the derivative of the square of the secant function is $2\tan(x)\sec^2(x)$ so we have$$\sec^2(x)=\dfrac d{dx}(2\tan(x)\sec^2(x))$$and now to manipulate the top part. So recognizing that we need to get this into $\frac d{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$ form, we have$$3e^{2x}\sin(x)+4e^{2x}\cos(x)=4e^{2x}\sin(x)-e^{2x}\sin(x)+2e^{2x}\cos(x)+2e^{2x}\cos(x)\\=\dfrac d{dx}(2e^{2x}\sin(x)+e^{2x}\cos(x))$$so now the integral is just$$-\int\dfrac{\frac d{dx}(2e^{2x}\sin(x)+e^{2x}\cos(x))}{\frac d{dx}(2\tan(x)\sec^2(x))}dx$$And since I know that$$\dfrac d{dx}\dfrac{f(x)}{g(x)}=\dfrac{\frac d{dx}f(x)}{\frac d{dx}g(x)}$$because of L'Hôpital's Rule, we have our antiderivative to be$$\dfrac{-(2e^{2x}\sin(x)+e^{2x}\cos(x))}{2\tan(x)\sec^2(x)}+c$$or just$$-e^{2x}\sin(x)\cos^2(x)\cot(x)-0.5e^{2x}\cos^3(x)\cot(x)+c$$

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My question

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Did I calculate the antiderivative correctly, or what could I do to caculate the antiderivative?

Answer 1

Without any integration by parts.

After the simplification of the denominator $$I=-\int e^{2 x} \cos ^2(x) (3 \sin (x)+4 \cos (x))\,dx$$

Linearizing the powers $$I=-\int e^{2x} \left(\frac{3 }{4}\sin (x)+\frac{3}{4} \sin (3 x)+3 \cos(x)+\cos (3 x)\right)\,dx$$ Use Euler representation of the sine and cosine functions to just face four exponentials with complex arguments which are more than simple to integrate.

Taking the real part of the result $$I=-\frac{1}{260} e^{2 x} (234 \sin (x)+90 \sin (3 x)+273 \cos (x)-5 \cos (3 x))$$

Answer 2

You calculated the denominator as $-\sec^2x$. So your monster integral is $$\int(-3e^{2x}\sin x\cos^2x-4e^{2x}\cos^3x)dx=\int e^{2x}(\cos^3x)'dx-4\int e^{2x}\cos^3xdx$$ By integration by parts for the first integral we have $$e^{2x}\cos^3x-6\int e^{2x}\cos^3xdx.$$ That is where I am disappointed. I expected cancellation. If you change $+4$ to $-2$, then the answer would be $e^{2x}\cos^3x.$

Answer 3

$$ \begin{aligned} \int \frac{3 e^{2 x} \sin x+4 e^{2 x} \cos x}{-\left(\sin ^2 x+\cos ^2 x+\tan ^2 x\right)} d x = & -\int \frac{3 e^{2 x} \sin x+4 e^{2 x} \cos x}{\sec ^2 x} d x \\ = & \int\left(-3e^{2 x} \sin x \cos ^2 x-4 e^{2 x} \cos ^3 x\right) d x \\ = & \int e^{2 x} d\left(\cos ^3 x\right)-4 \int e^{2 x} \cos ^3 x d x \\ = & e^{2 x} \cos ^3 x-2 \int e^{2 x} \cos ^3 x d x-4 \int e^{2 x} \cos ^3 x d x \\ = & e^{2 x} \cos ^3 x-6\int e^{2 x} \cos ^3 x d x \end{aligned} $$ For the last integral, we can use the formula: $\cos 3 x=4 \cos ^3 x-3 \cos x$ and get $$ \int e^{2 x} \cos ^3 x d x =\frac{1}{4}\left[\int e^{2 x} \cos 3 x d x+3 \int e^{2 x} \cos x d x\right] $$ Using integration by parts twice, we get a formula $$ \int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x) $$ and obtain $$ \int e^{2 x} \cos ^3 x d x =\frac{e^{2 x}}{260}\left( 15 \sin 3 x+10 \cos 3 x+39 \sin x+78 \cos x\right) $$ Hence we can conclude that $$\boxed{\quad \int \frac{3 e^{2 x} \sin x+4 e^{2 x} \cos x}{-\left(\sin ^2 x+\cos ^2 x+\tan ^2 x\right)} d x \\= e^{2 x} \cos ^3 x - \frac{3e^{2x}}{130}\left( 15 \sin 3 x+10 \cos 3 x+39 \sin x+78 \cos x\right) +C}$$