Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.2

Integrate the following functions with respect to x

Question 1.
(i) \( (x + 5)^6 \)
(ii) \( \frac{1}{(2-3x)^4} \)
(iii) \( \sqrt{3x+2} \)
Answer:
(i) To integrate \( (x + 5)^6 \), we use the formula \( \int (ax + b)^n dx = \frac{1}{a} \frac{(ax + b)^{n+1}}{n+1} + c \).
Here, \( a = 1 \), \( b = 5 \), \( n = 6 \).
\[ \int (x + 5)^6 dx = \frac{1}{1} \frac{(x + 5)^{6+1}}{6+1} + c \]
\[ = \frac{(x + 5)^7}{7} + c \]
(ii) To integrate \( \frac{1}{(2-3x)^4} \), we first rewrite it as \( (2-3x)^{-4} \).
Using the same formula \( \int (ax + b)^n dx = \frac{1}{a} \frac{(ax + b)^{n+1}}{n+1} + c \).
Here, \( a = -3 \), \( b = 2 \), \( n = -4 \).
\[ \int \frac{1}{(2-3x)^4} dx = \int (2-3x)^{-4} dx \]
\[ = \frac{1}{-3} \frac{(2-3x)^{-4+1}}{-4+1} + c \]
\[ = \frac{1}{-3} \frac{(2-3x)^{-3}}{-3} + c \]
\[ = \frac{1}{9} (2-3x)^{-3} + c \]
\[ = \frac{1}{9(2-3x)^3} + c \]
(iii) To integrate \( \sqrt{3x+2} \), we first rewrite it as \( (3x+2)^{\frac{1}{2}} \).
Using the formula \( \int (ax + b)^n dx = \frac{1}{a} \frac{(ax + b)^{n+1}}{n+1} + c \).
Here, \( a = 3 \), \( b = 2 \), \( n = \frac{1}{2} \).
\[ \int \sqrt{3x+2} dx = \int (3x+2)^{\frac{1}{2}} dx \]
\[ = \frac{1}{3} \frac{(3x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1} + c \]
\[ = \frac{1}{3} \frac{(3x+2)^{\frac{3}{2}}}{\frac{3}{2}} + c \]
\[ = \frac{1}{3} \times \frac{2}{3} (3x+2)^{\frac{3}{2}} + c \]
\[ = \frac{2}{9} (3x+2)^{\frac{3}{2}} + c \]
In simple words: We used a standard rule for integrating expressions like \( (ax+b)^n \). We raise the power by one and divide by the new power, also dividing by the 'a' part from inside the bracket. Remember to change roots and fractions into powers before integrating.

🎯 Exam Tip: Always remember the constant of integration \( + c \) when evaluating indefinite integrals. Also, correctly identify 'a' and 'n' in the formula to avoid calculation errors.

Question 2.
(i) \( \sin 3x \)
(ii) \( \cos(5-11x) \)
(iii) \( \text{cosec}^2(5x-7) \)
Answer:
(i) To integrate \( \sin 3x \), we use the formula \( \int \sin(ax+b) dx = -\frac{1}{a} \cos(ax+b) + c \).
Here, \( a = 3 \), \( b = 0 \).
\[ \int \sin 3x dx = -\frac{1}{3} \cos 3x + c \]
(ii) To integrate \( \cos(5-11x) \), we use the formula \( \int \cos(ax+b) dx = \frac{1}{a} \sin(ax+b) + c \).
Here, \( a = -11 \), \( b = 5 \).
\[ \int \cos(5-11x) dx = \frac{1}{-11} \sin(5-11x) + c \]
\[ = -\frac{1}{11} \sin(5-11x) + c \]
(iii) To integrate \( \text{cosec}^2(5x-7) \), we use the formula \( \int \text{cosec}^2(ax+b) dx = -\frac{1}{a} \cot(ax+b) + c \).
Here, \( a = 5 \), \( b = -7 \).
\[ \int \text{cosec}^2(5x-7) dx = -\frac{1}{5} \cot(5x-7) + c \]
In simple words: We used basic integration rules for trigonometric functions like sine, cosine, and cosecant squared. For these, we get new trigonometric functions, and we must remember to divide by the coefficient of 'x' and add the integration constant.

🎯 Exam Tip: Pay close attention to the sign changes when integrating sine and cosine functions. Also, ensure you correctly identify the 'a' coefficient, especially when it's negative.

Question 3.
(i) \( e^{3x-6} \)
(ii) \( e^{8-7x} \)
(iii) \( \frac{1}{6-4x} \)
Answer:
(i) To integrate \( e^{3x-6} \), we use the formula \( \int e^{ax+b} dx = \frac{1}{a} e^{ax+b} + c \).
Here, \( a = 3 \), \( b = -6 \).
\[ \int e^{3x-6} dx = \frac{1}{3} e^{3x-6} + c \]
(ii) To integrate \( e^{8-7x} \), we use the formula \( \int e^{ax+b} dx = \frac{1}{a} e^{ax+b} + c \).
Here, \( a = -7 \), \( b = 8 \).
\[ \int e^{8-7x} dx = \frac{1}{-7} e^{8-7x} + c \]
\[ = -\frac{1}{7} e^{8-7x} + c \]
(iii) To integrate \( \frac{1}{6-4x} \), we use the formula \( \int \frac{dx}{ax+b} = \frac{1}{a} \log|ax+b| + c \).
Here, \( a = -4 \), \( b = 6 \).
\[ \int \frac{dx}{6-4x} = \frac{1}{-4} \log|6-4x| + c \]
\[ = -\frac{1}{4} \log|6-4x| + c \]
In simple words: For expressions with 'e' to the power of a linear term, we keep the 'e' part the same and divide by the number multiplying 'x'. For fractions like \( \frac{1}{ax+b} \), the integral is related to the natural logarithm of the denominator, divided by the 'x' coefficient.

🎯 Exam Tip: Remember that the integral of \( \frac{1}{x} \) is \( \log|x| \). For \( \frac{1}{ax+b} \), the 'a' in the denominator is crucial. Always check the sign of the 'a' value carefully.

Question 4.
(i) \( \sec^2 \frac{x}{5} \)
(ii) \( \text{cosec}(5x+3) \cot(5x+3) \)
(iii) \( 30 \sec(2-15x) \tan(2-15x) \)
Answer:
(i) To integrate \( \sec^2 \frac{x}{5} \), we use the formula \( \int \sec^2(ax+b) dx = \frac{1}{a} \tan(ax+b) + c \).
Here, \( a = \frac{1}{5} \), \( b = 0 \).
\[ \int \sec^2 \frac{x}{5} dx = \frac{1}{\frac{1}{5}} \tan \frac{x}{5} + c \]
\[ = 5 \tan \frac{x}{5} + c \]
(ii) To integrate \( \text{cosec}(5x+3) \cot(5x+3) \), we use the formula \( \int \text{cosec}(ax+b) \cot(ax+b) dx = -\frac{1}{a} \text{cosec}(ax+b) + c \).
Here, \( a = 5 \), \( b = 3 \).
\[ \int \text{cosec}(5x+3) \cot(5x+3) dx = -\frac{1}{5} \text{cosec}(5x+3) + c \]
(iii) To integrate \( 30 \sec(2-15x) \tan(2-15x) \), we first take the constant 30 out.
Then we use the formula \( \int \sec(ax+b) \tan(ax+b) dx = \frac{1}{a} \sec(ax+b) + c \).
Here, \( a = -15 \), \( b = 2 \).
\[ \int 30 \sec(2-15x) \tan(2-15x) dx = 30 \times \frac{1}{-15} \sec(2-15x) + c \]
\[ = -2 \sec(2-15x) + c \]
In simple words: We used specific integration rules for different combinations of trigonometric functions. The integral of \( \sec^2 \) gives \( \tan \), and the integral of \( \text{cosec} \cot \) gives \( -\text{cosec} \). For \( \sec \tan \), the integral gives \( \sec \). Always divide by the coefficient of 'x' in the argument.

🎯 Exam Tip: Familiarize yourself with the common trigonometric integral formulas. Remember to handle any constants multiplying the function by bringing them out of the integral, and always divide by the coefficient of 'x'.

Question 5.
(i) \( \frac{1}{\sqrt{1-(4x)^2}} \)
(ii) \( \frac{1}{\sqrt{1-81x^2}} \)
(iii) \( \frac{1}{1+36x^2} \)
Answer:
(i) To integrate \( \frac{1}{\sqrt{1-(4x)^2}} \), we use the formula \( \int \frac{dx}{\sqrt{1-(ax+b)^2}} = \frac{1}{a} \sin^{-1}(ax+b) + c \).
Here, \( a = 4 \), \( b = 0 \).
\[ \int \frac{dx}{\sqrt{1-(4x)^2}} = \frac{1}{4} \sin^{-1}(4x) + c \]
(ii) To integrate \( \frac{1}{\sqrt{1-81x^2}} \), we rewrite \( 81x^2 \) as \( (9x)^2 \).
So, it becomes \( \int \frac{dx}{\sqrt{1-(9x)^2}} \).
Using the formula \( \int \frac{dx}{\sqrt{1-(ax+b)^2}} = \frac{1}{a} \sin^{-1}(ax+b) + c \).
Here, \( a = 9 \), \( b = 0 \).
\[ \int \frac{dx}{\sqrt{1-81x^2}} = \int \frac{dx}{\sqrt{1-(9x)^2}} \]
\[ = \frac{1}{9} \sin^{-1}(9x) + c \]
(iii) To integrate \( \frac{1}{1+36x^2} \), we rewrite \( 36x^2 \) as \( (6x)^2 \).
So, it becomes \( \int \frac{dx}{1+(6x)^2} \).
Using the formula \( \int \frac{dx}{1+(ax+b)^2} = \frac{1}{a} \tan^{-1}(ax+b) + c \).
Here, \( a = 6 \), \( b = 0 \).
\[ \int \frac{dx}{1+36x^2} = \int \frac{dx}{1+(6x)^2} \]
\[ = \frac{1}{6} \tan^{-1}(6x) + c \]
In simple words: These problems use special integration formulas that lead to inverse trigonometric functions like \( \sin^{-1} \) (arcsin) and \( \tan^{-1} \) (arctan). We need to make sure the term with 'x' inside the square root or in the denominator is in the form \( (ax)^2 \) to apply these rules correctly.

🎯 Exam Tip: Recognize the forms \( \frac{1}{\sqrt{1-u^2}} \) and \( \frac{1}{1+u^2} \) for \( \sin^{-1}u \) and \( \tan^{-1}u \) integrals, respectively. Ensure you correctly identify 'u' as \( (ax) \) and divide by 'a'.