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

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

  • $\begingroup$ This is not an integral to do by parts. $\endgroup$
    – user296602
    Dec 10, 2018 at 17:22
  • $\begingroup$ Your by-parts integration is wrong. $\endgroup$
    – user65203
    Dec 10, 2018 at 17:31
  • $\begingroup$ 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. $\endgroup$
    – JavaMan
    Dec 10, 2018 at 17:33
  • $\begingroup$ Please use mathjax to format your equations next time so we don't have to do it for you $\endgroup$
    – Jam
    Dec 10, 2018 at 17:51
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Dec 10, 2018 at 17:53

4 Answers 4


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

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

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

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



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,


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

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

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



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

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


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

  • 2
    $\begingroup$ 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. $\endgroup$
    – JavaMan
    Dec 10, 2018 at 17:44

By parts,

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

leads you about nowhere.


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