# Help to calculate the integral $\int 314^{\cos x} \; dx$

I stopped at the place highlighted in yellow how to find this integral ? $$\int (314)^{\large \cos x} \; dx$$

\begin{aligned}\int 314^{\cos x}\sin x\,\mathrm{d}x=&\int u\mathrm{d}v=uv-\int v\mathrm{d} u\\ &u=\sin x\quad\mathrm{d}u=\cos x\mathrm{d}x\\ &\mathrm{d}v=314^{\cos x}\mathrm{d}x\quad v=\int 314^{\cos x}\mathrm{d}x \end{aligned}

• This is not an integral to do by parts.
– user296602
Commented Dec 10, 2018 at 17:22
• Your by-parts integration is wrong.
– user65203
Commented Dec 10, 2018 at 17:31
• The integrals $\int a^{\cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = \cos x$ as indicated by the answer below. Commented Dec 10, 2018 at 17:33
• Please use mathjax to format your equations next time so we don't have to do it for you
– Jam
Commented Dec 10, 2018 at 17:51
• The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = \sin x$ and $dv = 314^{\cos x}dx$ is the exact opposite if that. Try $dv= \sin x$ so $v =-\cos x$ and $u=314{-v}$ and $\int 314^{\cos x} dx = \int u dv = \int 314^{-v} dv$. (ALthough switching the negatives so $v=\cos x$ and $\int - 314^{v}dv$ will be easier. Commented Dec 10, 2018 at 17:53

Hint: Substitute $$t=\cos(x)$$ then $$dt=-\sin(x)dx$$Then you will get $$-\int 314^t dt$$

• But its $\ dx$ how to complete the rest? Commented Dec 11, 2018 at 7:48
• it is $$dt=-\sin(x)dx$$ what do you mean with the rest? Commented Dec 11, 2018 at 8:33
• $$\int314^{\cos x}\ dx=-\int\dfrac{314^t\ dt}{\sin x}$$ right? Commented Dec 11, 2018 at 8:40
• I thought your integral is $$\int 314^{\cos(x)}\sin(x)dx$$ Commented Dec 11, 2018 at 8:42
• Where have you find this $$\int314^{\cos x}\sin x\ dx$$ Commented Dec 11, 2018 at 8:43

$$\int314^{\cos x}~dx$$

$$=\int e^{\ln314\cos x}~dx$$

$$=\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n}314\cos^{2n}x}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}314\cos^{2n+1}x}{(2n+1)!}dx$$

$$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}314\cos^{2n}x}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}314\cos^{2n+1}x}{(2n+1)!}dx$$

For $$n$$ is any natural number,

$$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$$

This result can be done by successive integration by parts.

For $$n$$ is any non-negative integer,

$$\int\cos^{2n+1}x~dx$$

$$=\int\cos^{2n}x~d(\sin x)$$

$$=\int(1-\sin^2x)^n~d(\sin x)$$

$$=\int\sum\limits_{k=0}^nC_k^n(-1)^k\sin^{2k}x~d(\sin x)$$

$$=\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$$

$$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}314\cos^{2n}x}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}314\cos^{2n+1}x}{(2n+1)!}dx$$

$$=x+\sum\limits_{n=1}^\infty\dfrac{x\ln^{2n}314}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}314\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}314\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$$

$$=\sum\limits_{n=0}^\infty\dfrac{x\ln^{2n}314}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}314\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}314\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$$

The whole point of substitution by parts is to substitute.

You are trying to find

$$\int 314^{\cos x} \sin x dx$$.

If you substitute $$u = \cos x$$ then $$314^{\cos x}$$ becomes $$314^u$$.

And $$du = -\sin x dx$$ so $$\sin x dx$$.

So $$\int 314^{\cos x} \sin x dx = \int -314^{u}du$$.

And as $$\sin a^x dx = \frac {a^x}{\ln a} + C$$

We have $$\int 314^{\cos x} \sin x dx = \int -314^{u}du= -\frac {314^u}{\ln 314} = -\frac {314^{\cos x}}{\ln 314}$$

• Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule. Commented Dec 10, 2018 at 17:44

By parts,

$$\int 314^{\cos x}\sin x\,dx=-314^{\cos x}\cos x-\log314\int 314^{\cos x}\sin^2x\,dx$$