Evaluation of Definite Integrals by Substitution

Definite Integral by Substitution as the limit of a sum is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of integration have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concept of definite Integrals by Substitution. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), thirteen questions have been asked on this concept, including one in 2016, one in 2017, one in 2018, two in 2022, five in 2021, and three in 2022.

Evaluation of Definite Integrals by Substitution

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y about x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

Substitution is one of the basic methods for calculating indefinite integrals. This technique transforms a complex integral into a simpler one by changing the variable of integration. It is especially useful for integrals involving composite functions where a direct integration approach is difficult.

We have already learned to find Indefinite Integration by using the substitution method. But in the case of definite integration, we also need to change the limits of integration 'a' and 'b'. If we substitute x = g(t), then g(t) must be continuous in the interval [a, b].

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b] and F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then $\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Let's see some examples to see how such questions are solved.

Example 1

Compute the integral $\int_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$

Let $\quad I=\int_{x=0}^{x=\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Divide numerator and denominator by $\cos ^2 x$

$
=\int_{x=0}^{x=\pi / 2} \frac{\sec ^2 x d x}{a^2+b^2 \tan ^2 x}
$

Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$

$
\therefore \quad I=\int_{t=0}^{t=\infty} \frac{d t}{a^2+b^2 t^2}
$
We find the new limits of integration $t=\tan x \Rightarrow t=0$ when $x=0$ and $t=\infty$ when $x=\pi / 2$$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{b^2} \int_0^{\infty} \frac{d t}{\left(\frac{a}{b}\right)^2+t^2}=\frac{1}{b^2} \cdot \frac{1}{a / b}\left[\tan ^{-1} \frac{b t}{a}\right]_0^{\infty} \\
& =\frac{1}{a b}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 a b}
\end{aligned}

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Example 1: The integral $\int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x$equals:

1) $e(4 e+1)$
2) $4 e^2-1$
3) ${ }^{e(4 e-1)}$
4) $e(2 e-1)$

Solution

$\begin{aligned} & \int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x \\ & \int_1^2 e^x\left(2 x^x+x^x \log _e x\right) d x \\ & \int_1^2 e^x(\underbrace{x^x}_{f(x)}+\underbrace{x^x\left(1+\log _e x\right)}_{f^{\prime}(x)}) d x \\ & \left(e^x \cdot x^x\right)_1^2=4 e^2-e\end{aligned}$

Hence, the answer is the option 3.

Example 2: Let f be a twice differentiable function defined on R such that $f(0)=1, f^{\prime}(0)=2$ and $f^{\prime}(x) \neq 0$ for all$x \in R$. If $\left|\begin{array}{cc}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ for all $x \in R$ then the value of $f(1)$ lies in the interval:

1) $(0,3)$
2) $(9,12)$
3) $(6,9)$
4) $(3,6)$

Solution

$\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}=0$

$\begin{aligned} & f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\ & \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\ & \ln \left(f^{\prime}(x)\right)=\ln f(x)+\ln c\end{aligned}$

$\begin{aligned} & f^{\prime}(x)=c f(x) \\ & \frac{f^{\prime}(x)}{f(x)}=c \\ & \ln f(x)=c x+k\end{aligned}$

$\begin{aligned} & f(x)=k e^{c x} \\ & f(0)=1=k \\ & f^{\prime}(0)=c=2 \\ & f(x)=e^{2 x} \\ & f(1)=e^2 \in(6,9)\end{aligned}$

Hence, the answer is the option 3.

Example 3: For x>0 , if $f(x)=\int_1^x \frac{\log _e t}{1+t} d t$, then $f(e)+f\left(\frac{1}{e}\right)$ is equal to

1) 1

2) -1

3)

4) 0

Solution

$\begin{aligned} & f(x)=\int_1^x \frac{\log _e t}{(1+t)} d t \\ & f\left(\frac{1}{x}\right)=\int_1^{1 / x} \frac{\log _e t}{1+t} d t \\ & \operatorname{let} t=\frac{1}{y}\end{aligned}$

$\begin{aligned} & =\int_1^x \frac{\log _e y}{1+y} \cdot \frac{y}{y^2} d y \\ & =\int_1^x \frac{\log _e y}{y(1+y)} d y\end{aligned}$

Hence,

$f(x)+f\left(\frac{1}{x}\right)=\int_1^x \frac{(1+t) \log _c t}{t(1+t)} d t=\int_1^x \frac{\log _e t}{t} d t$

put $u=\log _{\mathrm{c}} t \Rightarrow \frac{d u}{d t}=\frac{1}{t}$

$\begin{gathered}=\frac{1}{2} \log _{\mathrm{e}}^2(\mathrm{x}) \\ \text { so } f(\mathrm{e})+f\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{2}\end{gathered}$

Hence, the answer is the option 3.

Example 4: Let $f(x)=\int_0^x e^t f(t) d t+e^x$ be a differentiable function for all . Then f(x) equals:

1) (correct)

$$
2 e^{\left(\mathrm{e}^x-1\right)}-1
$$

2) $e^{\left(e^x-1\right)}$
3) $e^{e^x}-1$
4) $2 e^{e^x}-1$

Solution

$f(\mathrm{x})=\int_0^{\mathrm{x}} \mathrm{e}^{\mathrm{t}} f(\mathrm{t}) \mathrm{dt}+\mathrm{e}^{\mathrm{x}} \Rightarrow f(0)=1$

differentiating with respect to x

$\begin{aligned} & f(x)=e^x f(x)+e^x \\ & f^{\prime}(x)=e^x(f(x)+1) \\ & \int_0^x \frac{f^{\prime}(x)}{f(x)+1} d x=\int_0^x e^x d x \\ & \left.\ln (f(x)+1)\right|_0 ^x=\left.e^x\right|_0 ^x\end{aligned}$

$\begin{aligned} & n(f(x)+1)-\ln (f(0)+1)=e^2-1 \\ & \ln \left(\frac{f(x)+1}{2}\right)-e^x-1 \quad(\text { as } f(0)=1) \\ & f(x)=2 e^{\left(c^2-1\right)}-1\end{aligned}$

Hence, the answer is the option 1.

Example 5: If $I=\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x$ . Then find the value of

1) $2 \ln 2$
2) $\ln 2$
3) $2 \ln \frac{\pi}{2}$
4) $\ln \frac{\pi}{2}$

Solution

Let $u=1+\sin x$

$d u=\cos x d x$

When $x=0, u=1+\sin (0)=1$
When $x=\pi / 2, \quad u=1+\sin \left(\frac{\pi}{2}\right)=2$

Then

$\int_0^{\pi / 2} \frac{\cos x}{1+\sin x}=\int_1^2 u^{-1} d u=\ln |u|_1^2=[\ln 2-\ln 1]-\ln 2$

Hence, the answer is the option 2.

Summary

Definite integral by Substitution is a useful technique in integration that allows us to simplify and solve complex integrals. Mastery of integration is essential for progressing in algebra, calculus, and applied mathematics, offering valuable tools for both theoretical and practical problem-solving.