RD Sharma Class 12 Exercise 19.4(a) Definite Integrals Solutions Maths

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

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:19 PM IST

The students of class 12 get the help of their teachers to clarify their doubts. Yet this cannot take place 24 x 7, especially when the students are at their homes trying out the homework by themselves. Definite Integrals is one such chapter where the help of proper guidance is essential for the students. The RD Sharma Class 12th Exercise 19.4(a) solution book plays a significant role in such cases.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise

Definite Integrals Ex 19.4(a),

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

Frequently Asked Questions (FAQs)

1. Which is the most recommended solution book for the class 12 students to clarify their queries regarding the Definite Integrals concept?

The RD Sharma Class 12th Exercise 19.4(a) book is the most recommended reference material for the students who have doubts about the Definite Integrals concept.

2. Are the RD Sharma books available for free?

The RD Sharma books are available for free of cost at the Career 360 website, making them accessible for anyone who needs it.

3. How many parts of Class 12, chapter 19, Exercise 4 are available in the RD Sharma collection?

There are two parts, Ex 19.4(a) and Ex 19.4(b), in the fourth exercise of the 19 chapters in mathematics. The solutions for the first part are given in the RD Sharma Class 12th Exercise 19.4(a) book.

4. Is it possible to download the RD Sharma solution books?

Yes, the RD Sharma solution books available on the Career 360 website come with an option to be downloaded. Therefore, the students can refer to the solutions whenever required.

5. Can the CBSE board students refer to the RD Sharma books to clarify their doubts?

As the RD Sharma books follow the NCERT pattern, the CBSE Students can use it to understand the concept deeply.