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

Question

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

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

Answer 1

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

Answer 2

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

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

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

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

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,

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

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

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

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

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

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

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

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

Answer 3

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

Answer 4

By parts,

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

leads you about nowhere.