Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.1

Integrate the Following With Respect to x:

Question 1. Evaluate the following integrals:
(i) \( x^{11} \)
(ii) \( \frac{1}{x^{7}} \)
(iii) \( \sqrt[3]{x^{4}} \)
(iv) \( (x^{5})^{1/8} \)
Answer:
(i) To integrate \( x^{11} \):
\( \int x^{11} \,dx \)
\( = \frac{x^{11+1}}{11+1} + C \)
\( = \frac{x^{12}}{12} + C \)
When integrating power functions, we increase the power by one and divide by the new power.
(ii) To integrate \( \frac{1}{x^{7}} \):
\( \int \frac{1}{x^{7}} \,dx \)
\( = \int x^{-7} \,dx \)
\( = \frac{x^{-7+1}}{-7+1} + C \)
\( = \frac{x^{-6}}{-6} + C \)
\( = -\frac{1}{6x^{6}} + C \)
First, rewrite the fraction with a negative exponent to use the power rule for integration.
(iii) To integrate \( \sqrt[3]{x^{4}} \):
\( \int \sqrt[3]{x^{4}} \,dx \)
\( = \int x^{4/3} \,dx \)
\( = \frac{x^{4/3+1}}{4/3+1} + C \)
\( = \frac{x^{7/3}}{7/3} + C \)
\( = \frac{3}{7} x^{7/3} + C \)
Convert the radical expression into a fractional exponent before applying the power rule of integration.
(iv) To integrate \( (x^{5})^{1/8} \):
\( \int (x^{5})^{1/8} \,dx \)
\( = \int x^{5/8} \,dx \)
\( = \frac{x^{5/8+1}}{5/8+1} + C \)
\( = \frac{x^{13/8}}{13/8} + C \)
\( = \frac{8}{13} x^{13/8} + C \)
Simplify the exponent first by multiplying the powers, then apply the standard integration power rule.
In simple words: For each expression, add one to the power and then divide by that new power. Remember to always add 'C' at the end for indefinite integrals.

🎯 Exam Tip: Always convert roots and fractions with variables in the denominator to exponent form (like \( x^n \)) before integrating. Remember that \( \int x^n \,dx = \frac{x^{n+1}}{n+1} + C \).

Question 2. Integrate the following expressions with respect to x:
(i) \( \frac{1}{\sin ^{2}x} \)
(ii) \( \frac{\tan x}{\cos x} \)
(iii) \( \frac{\cos x}{\sin ^{2} x} \)
(iv) \( \frac{1}{\cos ^{2} x} \)
Answer:
(i) To integrate \( \frac{1}{\sin ^{2}x} \):
\( \int \frac{1}{\sin ^{2}x} \,dx \)
\( = \int \operatorname{cosec}^{2} x \,dx \)
\( = -\cot x + C \)
We use the trigonometric identity \( \frac{1}{\sin x} = \operatorname{cosec} x \) to simplify the expression before integrating.
(ii) To integrate \( \frac{\tan x}{\cos x} \):
\( \int \frac{\tan x}{\cos x} \,dx \)
\( = \int \sec x \tan x \,dx \)
\( = \sec x + C \)
We rewrite the expression using \( \frac{1}{\cos x} = \sec x \) to match a standard integral form.
(iii) To integrate \( \frac{\cos x}{\sin ^{2} x} \):
\( \int \frac{\cos x}{\sin ^{2} x} \,dx \)
\( = \int \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} \,dx \)
\( = \int \operatorname{cosec} x \cot x \,dx \)
\( = -\operatorname{cosec} x + C \)
Break the fraction into two parts, \( \frac{1}{\sin x} \) and \( \frac{\cos x}{\sin x} \), to simplify it to known trigonometric identities.
(iv) To integrate \( \frac{1}{\cos ^{2} x} \):
\( \int \frac{1}{\cos ^{2} x} \,dx \)
\( = \int \sec^{2} x \,dx \)
\( = \tan x + C \)
This integral directly uses the identity \( \frac{1}{\cos x} = \sec x \) and a common integral formula.
In simple words: For these problems, first change the given expression into a simpler trigonometric form using identities. Then, use the standard integration rules for those trigonometric functions.

🎯 Exam Tip: Memorize common trigonometric identities and their corresponding integral formulas (e.g., \( \int \operatorname{cosec}^2 x \,dx = -\cot x + C \)) to quickly solve these types of questions.

Question 3. Integrate the following expressions with respect to x:
(i) \( 123 \)
(ii) \( \frac{x^{24}}{x^{25}} \)
(iii) \( e^x \)
Answer:
(i) To integrate \( 123 \):
\( \int 123 \,dx \)
\( = 123 \int 1 \,dx \)
\( = 123x + C \)
Integrating a constant simply means multiplying the constant by the variable of integration.
(ii) To integrate \( \frac{x^{24}}{x^{25}} \):
\( \int \frac{x^{24}}{x^{25}} \,dx \)
\( = \int \frac{1}{x} \,dx \)
\( = \log |x| + C \)
First simplify the expression using exponent rules, then recall the specific integral for \( \frac{1}{x} \).
(iii) To integrate \( e^x \):
\( \int e^x \,dx \)
\( = e^x + C \)
The exponential function \( e^x \) is unique because its integral is also \( e^x \).
In simple words: Integrate each part separately. A number becomes "number times x", \( \frac{1}{x} \) becomes \( \log |x| \), and \( e^x \) stays \( e^x \). Don't forget the 'C'.

🎯 Exam Tip: Remember the special case for integrating \( \frac{1}{x} \), which is \( \log |x| \), not using the power rule \( \frac{x^{-1+1}}{-1+1} \), which would lead to division by zero.

Question 4. Integrate the following expressions with respect to x:
(i) \( (1 + x^2)^{-1} \)
(ii) \( (1 - x^2)^{-1/2} \)
Answer:
(i) To integrate \( (1 + x^2)^{-1} \):
\( \int (1 + x^2)^{-1} \,dx \)
\( = \int \frac{1}{1 + x^2} \,dx \)
\( = \tan^{-1} x + C \)
This integral is a direct application of the standard formula for inverse tangent. It's important to recognize these forms.
(ii) To integrate \( (1 - x^2)^{-1/2} \):
\( \int (1 - x^2)^{-1/2} \,dx \)
\( = \int \frac{1}{\sqrt{1 - x^2}} \,dx \)
\( = \sin^{-1} x + C \)
This integral is a direct application of the standard formula for inverse sine. Recognizing this form saves time.
In simple words: These are special integrals that give inverse trigonometric functions. \( \frac{1}{1+x^2} \) integrates to \( \tan^{-1} x \), and \( \frac{1}{\sqrt{1-x^2}} \) integrates to \( \sin^{-1} x \).

🎯 Exam Tip: Recognize standard integral forms that lead to inverse trigonometric functions. Knowing these specific formulas is crucial for quick and accurate solutions.