¿How do I integrate $e^\sqrt{x}$, using the substitution rule?

You're going to want to start with a $u$-substitution $u=\sqrt{x}$. Notice then that $u^2=x$, so $2udu=dx$. Then $$ \int e^\sqrt{x}dx=2\int e^u u\ du. $$

You can then proceed by integration by parts.

To get started, you can set $w=u$ and $dv=e^u\ du$, and use the fact that $\int w\ dv= wv-\int v\ dw$. Lastly, don't forget to substitute back to get the integral in terms of $x$.

Continuing, $dw=du$ and $v=e^u$. Then $$ \int w\ dv= wv-\int v\ dw = e^u u-\int e^u\ du=e^u u-e^u. $$ So $$ \int e^\sqrt{x}dx=2\int e^u u\ du=2e^uu-2e^u+C=2e^\sqrt{x}(\sqrt{x}-1)+C $$ after back substituting $u=\sqrt{x}$ and adding a possible constant term.

You can integrate $e^\sqrt{x}$ by substituting any single variable in place of $\sqrt{x}$.

Since, $e^\sqrt{x}$ does not look like $e^{x}$ due to the presence of square root, therefore it can not be integrated by normal methods.

$\int e^\sqrt{x} \,dx\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(1)$

To integrate $e^\sqrt{x}$, you should substitute any variable, say $t$ in place of $\sqrt{x}.$

Now, $e^\sqrt{x}$ becomes $e^t$ which is in the form of $e^x$, so it can be integrated easily.

But, when you substitute $t=$$\sqrt{x}$, then equation $(1)$ becomes :

$\int e^t \,dx$

and $t$ can not be integrated with respect to $x.$

So, replace $dx$ by evaluating its value as –

$t=\sqrt{x}$

$t^2=x$

On differentiation

$2tdt=dx$

Thus, $\int e^\sqrt{x}\,dx=\int e^t \cdot 2t\,dt$

$=2\int e^t \cdot t\,dt \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ...(2)$

Now, you have to integrate a product of two functions, so you need the formula for integration by parts.

$\int u\,dv=uv - \int v\,du$

Here, $u$ and $v$ should be assumed by following the order of ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric and Exponential) from left to right.

For example : If you have a product of two terms, say $x \cdot sec^2 x \,dx,$ then according to the order of ILATE, algebraic term $(x)$ comes before the trigonometric term $(sec^2 x).$

Therefore, you should consider the algebraic term $(x)$ as $u$ and the trigonometric term $(sec^2 x\,dx)$ as $dv$.

Now, going back to the original product i.e., $e^t \cdot t\,dt$, we find that $t$ is an algebraic term which comes before the exponential term, $e^t.$

So, let $u=t$ and $dv= e^t\,dt$

Now, you need the values of $v$ and $du$ for the formula.

$\int u\,dv=uv–\int v \,du$

Therefore, differentiate $u=t$ and integrate $dv=e^t\,dt$

$du=dt$ and $v=e^t$

Put all these values in the formula –

$\int e^t \cdot t \,dt=t \cdot e^t – \int e^t\,dt$

$=t \cdot e^t – e^t$

Now, from equation (2)

$\int e^\sqrt{x}\,dx = 2 \int e^t \cdot t \,dt$

$=2 (t \cdot e^t – e^t )+c$

Take $e^t$ out of the parenthesis

$=2 e^t (t–1)+c$

Put the value of $t$ to get the answer in terms of $x.$

So, $\int e^\sqrt{x}\,dx = 2 e^\sqrt{x} (\sqrt{x}–1)+c$

Here, $c$ is the constant.

This, is the integral of $\int e^\sqrt{x} \,dx.$

You can also integrate different types of functions using substitution method, which otherwise can’t be integrated by normal methods.

Let’s integrate $cos^3 x$ to understand the substitution rule better.

$\int cos^3 x\,dx$

$cos^3 x$ can not be integrated directly. So, we need to factorize it as :

$\int cos^2 x \cdot cos x\,dx$

Now, if we substitute $u=cos x$

then on differentiation, we get :

$du=–sin x\,dx$

But, we don’t have $–sin x$ in both of the factors .

i.e., $cos^2 x, cos x$

So, we can’t substitute $u=cos x$

For substitution, we should have both the functions, $sin x$ and $cos x$

$\because\, sin^2 x + cos^2 x = 1$

Subtracting $sin^2 x$ from both the sides

$\Rightarrow sin^2 x + cos^2 x – sin^2 x = 1 – sin^2 x$

$\Rightarrow sin^2 x – sin^2 x + cos^2 x = 1 – sin^2 x$

$cos^2x = 1 – sin^2 x$

So, we can replace $cos^2 x$ by $1 – sin^2 x$

Then $\int cos^2 x \cdot cos x\,dx$ becomes, $\int (1–sin^2 x) \cdot cos x\, dx$

Now, we can substitute $u=sin x$

Thus, $\int (1–sin^2 x)\cdot cos x \,dx$ becomes,

$\int (1–u^2) cos x \,dx$

Differentiate $u=sin x$

$du =cos x \,dx$

So, $\int (1–u^2) cos x \,dx$

$= \int (1–u^2) du$

Now, it is easy to integrate

$=\int 1 \cdot du – \int u^2 \,du$

$=u –\frac{u^{2+1}}{2+1} +c$

$=u –\frac{u^3}{3} +c$

put the value of $u $

Then, $\int cos^3 x \,dx = sin x – \frac{sin^3 x}{3} +c$

or

$\int cos^3 x \,dx = sin x(1 – \frac{sin^2 x}{3}) +c $