RD Sharma Class 12 Exercise 19.4(a) Definite Integrals Solutions Maths - Download PDF Free Online

Definite Integrals Exercise 19.4 (a) Question 1

Answer : $\pi$
Given : $\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} d x$
Hint : Use the formula of $\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x$
Solution : $I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} d x-------(1)$
$\begin{aligned} &\int_{0}^{2 \pi} \frac{e^{\sin (2 \pi-x)}}{e^{\sin (2 \pi-x)}+e^{-\sin (2 \pi-x)}} d x \\ &\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x \end{aligned}$
$I=\int_{0}^{2 \pi} \frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}} d x \quad[\sin (2 \pi-x)=-\sin x]---------(2)$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{2 \pi} \frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}} d x+\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{-\sin x}+e^{\sin x}} d x \\ &2 I=\int_{0}^{2 \pi} \frac{e^{-\sin x}+e^{\sin x}}{e^{-\sin x}+e^{\sin x}} d x \\ &2 I=\int_{0}^{2 \pi} d x \end{aligned}$
$2I=2\pi$
$I=\pi$

Definite Integrals Exercise 19.4 (a) Question 2

Answer : 0
Given : $\int_{0}^{2 \pi} \log (\sec x+\tan x) d x$
Hint : You must know about the trignometric identities
Solution : $I=\int_{0}^{2 \pi} \log (\sec x+\tan x) d x-------(1)$
We know that
$\begin{aligned} &\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x \\ &I=\int_{0}^{2 \pi} \log (\sec (2 \pi-x)+\tan (2 \pi-x)) d x \\ &I=\int_{0}^{2 \pi} \log (\sec x-\tan x) d x \quad-------(2)[\sec (2 \pi-x)=\sec x, \tan (2 \pi-x)=-\tan x] \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{2 \pi} \log (\sec x+\tan x)+\log (\sec x-\tan x) d x \\ &2 I=\int_{0}^{2 \pi} \log (1) d x\left[\tan ^{2} x+1=\sec ^{2} x\right] \\ &2 I=0[\log (1)=0] \\ &I=0 \end{aligned}$

Definite Integrals Exercise 19.4 Question 3

Answer : $\frac{\pi}{12}$
Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x$
Hint : Use the formula of $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x \quad-----(1)$
$\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\tan \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)+\sqrt{\cot \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan \left(\frac{\pi}{2}-x\right)}}{\tan \left(\frac{\pi}{2}-x\right)}+\sqrt{\cot \left(\frac{\pi}{2}-x\right)} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x \quad------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x+\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}$$\begin{aligned} &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &2 I=\frac{\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 4

Answer : $\frac{\pi}{12}$
Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Hint : Use the formula of $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \quad-----(1)$
$\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sin \left(\frac{\pi}{2}-x\right)+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} d x------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x+\sqrt{\cos x}}} d x+\frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}$
$\begin{aligned} &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &2 I=\frac{\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 5
Answer : $1-\frac{\pi}{4}$

Given : $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}} d x$
Hint : Use the formula of $f(x)$ when $f(x)$ is odd and even
Solution : $I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}} d x \quad-----(1)$
We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$\begin{aligned} &I=\int_{\frac{\pi}{-4}}^{\frac{\pi}{4}} \frac{\tan ^{2}(-x)}{1+e^{-x}} d x \\ &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{-x}} d x------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}}+\frac{\tan ^{2} x}{1+e^{-x}} d x \\ &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x+e^{x} \tan ^{2} x}{1+e^{x}} d x \\ &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \tan ^{2} x d x \end{aligned}$
We know $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$ when $f(x)$ is even
$\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \tan ^{2} x d x \\ &\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \quad\left[\sec ^{2} x-1=\tan ^{2} x\right] \end{aligned}$
$\begin{aligned} &I=(\tan x-x)_{0}^{\frac{\pi}{4}} \\ &I=\left(1-\frac{\pi}{4}\right)-(0-0) \\ &=1-\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 6

Answer : a
Given : $\int_{-a}^{a} \frac{1}{1+a^{x}} d x, a>0$
Hint : Use the formula of $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$
Solution : $I=\int_{-a}^{a} \frac{1}{1+a^{x}} d x \quad-----(1)$
We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$\begin{aligned} &I=\int_{-a}^{a} \frac{1}{1+a^{-x}} d x \\ &I=\int_{-a}^{a} \frac{a^{x}}{a^{x}+1} d x-------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} 2 I &=\int_{-a}^{a} \frac{1+a^{x}}{a^{x}+1} d x \\ 2 I &=\int_{-a}^{a} 1 . d x \end{aligned}$
$\begin{aligned} &2 I=(x)_{-a}^{a} \\ &2 I=2 a \\ &I=a \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 7

Answer : $\frac{\pi}{3}$
Given : $\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}} d x$
Hint : You must know about the definite integrals
Solution : $I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}} d x \quad-----(1)$
$\begin{aligned} &I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{-\tan x}} d x------(2) \\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}}+\frac{e^{\tan x}}{1+e^{\tan x}} d x \\ &2 I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} 1 \cdot d x \\ &2 I=\frac{2 \pi}{3} \\ &I=\frac{\pi}{3} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 8

Answer : $\frac{\pi}{4}$
Given : $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{x}} d x$
Hint : the formula of a of $\int_{a}^{b} f(x) d x$ and then add the two equations
Solution : $I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{x}} d x \quad-----(1)$
$\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2}(-x)}{1+e^{-x}} d x \\ &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{-x}} d x \end{aligned}$
$\begin{aligned} &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\ &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{e^{x} \cos ^{2} x}{e^{x}+1} d x \quad-----(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x+e^{x} \cos ^{2} x}{1+e^{x}} d x \\ &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x d x \end{aligned}$
$\begin{aligned} &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\left[\cos 2 x=2 \cos ^{2} x-1\right] \\ &I=\frac{1}{4}\left(x+\frac{\sin 2 x}{2}\right)_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\ &I=\frac{1}{4}\left[\left(\frac{\pi}{2}+0\right)-\left(\frac{-\pi}{2}+0\right)\right] \end{aligned}$
$\begin{aligned} &I=\frac{1}{4}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 9

Answer : 2
Given : $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}+1}{\cos ^{2} x} d x$
Hint : You must know about the concept of odd and even of $f(x)$
Solution : $I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}+1}{\cos ^{2} x} d x$
$\begin{aligned} &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x}+\frac{1}{\cos ^{2} x} d x \\ &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x}+\sec ^{2} x d x \end{aligned}$
Now, $\frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x}$ is odd and $\sec^{2}x$ is even
$\begin{aligned} &I=0+2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\ &{\left[\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { even and } 0, \text { odd }\right]} \\ &I=2(\tan x)_{0}^{\frac{\pi}{4}} \\ &I=2 \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 10

Answer : $\frac{b-a}{2}$
Given : $\int_{a}^{b} \frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} d x, n \in N, n \geq 2$
Hint : Apply the formula $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Solution : $I=\int_{a}^{b} \frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} d x----(1)$

$\\I=\int_{a}^{b} \frac{(a+b-x)^{\frac{1}{n}}}{(a+b-x)^{\frac{1}{n}}+x^{\frac{1}{n}}} d x\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]------(2)$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{a}^{b} \frac{(a+b-x)^{\frac{1}{n}}}{(a+b-x)^{\frac{1}{n}}+x^{\frac{1}{n}}}+\frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} d x \\ &2 I=\int_{a}^{b} \frac{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} \end{aligned}$
$\begin{aligned} &2 I=\int_{a}^{b} d x \\ &2 I=(x)_{a}^{b} \\ &2 I=b-a \\ &I=\frac{b-a}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 11

Answer : $\frac{-\pi}{2} \log 2$
Given : $\int_{0}^{\frac{\pi}{2}}(2 \log \cos x-\log \sin 2 x) d x$
Hint : Do integration by parts and you must know properties of log
Solution : $I=\int_{0}^{\frac{\pi}{2}}(2 \log \cos x-\log \sin 2 x) d x$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}}\left(\log \cos ^{2} x-\log \sin 2 x\right) d x \\ &I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{\cos ^{2} x}{\sin 2 x}\right) d x\left(\log a-\log b=\log \frac{a}{b}\right) \\ &I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{\cos x}{2 \sin x}\right) d x(\sin 2 x=2 \sin x \cos x) \end{aligned}$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}}(\log \cos x-\log \sin x-\log 2) d x \\ &I=-\int_{0}^{\frac{\pi}{2}} \log 2 d x\left[I=\int_{0}^{\frac{\pi}{2}} \log \cos x d x=\int_{0}^{\frac{\pi}{2}} \log \sin x d x\right] \\ &I=\frac{-\pi}{2} \log 2 \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 12

Answer : $\frac{a}{2}$
Given : $\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$
Hint : Use the formula of $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Solution : $I=\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x-------(1)$
$I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x-------(2)\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x \\ &2 I=\int_{0}^{a} 1 . d x \\ &2 I=(x)_{0}^{a} \end{aligned}$
$\begin{aligned} &2 I=(x)_{0}^{a} \\ &2 I=a \\ &I=\frac{a}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 13

Answer : $\frac{5}{2}$
Given : $\int_{0}^{5} \frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}} d x$
Hint : Use the formula of $\int_{0}^{a} f(x) d x$
Solution : $I=\int_{0}^{5} \frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}} d x-------(1)$
$\begin{aligned} &I=\int_{0}^{5} \frac{\sqrt[4]{(5-x)+4}}{\sqrt[4]{(5-x)+4}+\sqrt[4]{9-(5-x)}} d x\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right) \\ &I=\int_{0}^{5} \frac{\sqrt[4]{9-x}}{\sqrt[4]{9-x}+\sqrt[4]{4+x}} d x------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{5} \frac{\sqrt[4]{x+4}+\sqrt[4]{9-x}}{\sqrt[4]{9-x}+\sqrt[4]{4+x}} d x \\ &2 I=\int_{0}^{5} 1 \cdot d x \\ &2 I=(x)_{0}^{5} \end{aligned}$
$\begin{aligned} &2 I=5 \\ &I=\frac{5}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 14

Answer : $\frac{7}{2}$
Given : $\int_{0}^{7} \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}} d x$
Hint : Use the formula of $\int_{0}^{a} f(x) d x$
Solution : $I=\int_{0}^{7} \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}} d x-------(1)$
$I=\int_{0}^{7} \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}} d x\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)----(2)$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{7} \frac{\sqrt[3]{x}+\sqrt[3]{7-x}}{\sqrt[3]{x}+\sqrt[3]{7-x}} d x \\ &2 I=\int_{0}^{7} 1 . d x \end{aligned}$
$\begin{aligned} \\ &2 I=(x)_{0}^{7} \\ &2 I=7 \\ &I=\frac{7}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (a) Question 15

Answer : $\frac{\pi}{12}$
Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x$
Hint : Use the formula of $\int_{a}^{b} f(x) d x$
Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x$
$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x----(1)\left(\tan x=\frac{\sin x}{\cos x}\right)$
We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x-------(2) \end{aligned}$
Add (1) and (2)
$\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 . d x \\ &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{aligned}$
$\begin{aligned} &2 I=\frac{2 \pi-\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}$

The syllabus of class 12, mathematics, chapter 9, Definite Integrals, is where the students require help during their starting stage. This fourth exercise consists of two parts, part 1 and part 2 (Ex 19.4(a) and Ex 19.4(b)). Exercise 19.4(a) consists of sixteen sums in total. The concept in these sums is evaluating the Definite integrals. This portion consists of basics that will help the students solve the sums in the second part of this exercise. To clarify their doubts in the first portion of this exercise, the students can use the RD Sharma Class 12th Exercise 19.4(a) solution book.

All the solutions in the RD Sharma books are contributed by a group of experts who have good experience in the field of education and mathematics. Numerous ways in which a solution can be derived are given. These methods are provided, taking into consideration the understanding capability of the students. Therefore, every student understands the concepts using the RD Sharma Class 12 Chapter 19 Exercise 19.4(a). the students can adapt to the method that they find easy and learn the mathematical concepts effortlessly.

The Class 12 RD Sharma Chapter 19 Exercise 19.4(a) Solution material consists of numerous practice questions that help the students attain better practice before exams. Moreover, it lets them face any problems in the Definite Integrals concept easily. As the RD Sharma books are based on the NCERT syllabus, the CBSE board students can use them for their reference without any hesitation.

The best part is that the RD Sharma Class 12th Exercise 19.4(a) book is currently available for free access at the Career 360 website. Students can download the resource material and use it while doing homework, assignments, and exam preparation. This has made every student easily access the best set of solutions.

Many students have attained good marks by preparing with RD Sharma Solutions Definite Integrals Ex 19.4(a) for their exams. As the questions for the public exam are taken from this book, students who practice with these questions score more than their fellow students.

RD Sharma Chapter wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable