Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.8

Integrate the Following With Respect to X:

Question 1.
(i) \( e^{ax} \cos bx \)
(ii) \( e^{2x} \sin x \)
(iii) \( e^{-x} \cos 2x \)
Answer:
(i) To integrate \( e^{ax} \cos bx \) with respect to \( x \), we use the standard formula for this type of integral. The formula states that the integral of \( e^{ax} \cos bx \) is \( \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C \). This formula helps to solve many problems involving exponential and trigonometric functions together.
\( \int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} [a \cos bx + b \sin bx] + C \)
(ii) To integrate \( e^{2x} \sin x \), we apply the general formula for integrals of the form \( \int e^{ax} \sin bx \, dx \). This formula is \( \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \). For this specific problem, we have \( a = 2 \) and \( b = 1 \). Substituting these values into the formula gives us the result. The constant of integration, \( C \), is always added for indefinite integrals.
Here \( a = 2, b = 1 \)
\( \int e^{2x} \sin x \, dx = \frac{e^{2x}}{2^2 + 1^2} [2 \sin x - 1 \cos x] + C \)
\( = \frac{e^{2x}}{4 + 1} [2 \sin x - \cos x] + C \)
\( = \frac{e^{2x}}{5} [2 \sin x - \cos x] + C \)
(iii) To integrate \( e^{-x} \cos 2x \), we use the standard formula for integrals of the form \( \int e^{ax} \cos bx \, dx \). The formula is \( \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C \). In this case, \( a \) is -1 and \( b \) is 2. We substitute these values into the formula and simplify the expression to get the final integral, ensuring all signs are correct.
Here \( a = -1, b = 2 \)
\( \int e^{-x} \cos 2x \, dx = \frac{e^{-x}}{(-1)^2 + 2^2} [(-1) \cos 2x + (2) \sin 2x] + C \)
\( = \frac{e^{-x}}{1 + 4} [2 \sin 2x - \cos 2x] + C \)
\( = \frac{e^{-x}}{5} [2 \sin 2x - \cos 2x] + C \)
In simple words: For these integrals, we use specific formulas that combine exponential and trigonometric parts. We identify the 'a' and 'b' values from the question and put them into the correct formula to find the answer.

🎯 Exam Tip: Remember the two key formulas for integrating \( e^{ax} \sin bx \) and \( e^{ax} \cos bx \), paying close attention to the signs and coefficients (a and b) in each term.

Question 2.
(i) \( e^{-3x} \sin 2x \)
(ii) \( e^{-4x} \sin 2x \)
(iii) \( e^{-3x} \cos x \)
Answer:
(i) To find the integral of \( e^{-3x} \sin 2x \), we use the general formula for integrating expressions of the form \( \int e^{ax} \sin bx \, dx \). This formula is \( \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \). In this case, \( a \) is -3 and \( b \) is 2. We carefully substitute these values into the formula and perform the calculations to arrive at the solution.
Here \( a = -3, b = 2 \)
\( \int e^{-3x} \sin 2x \, dx = \frac{e^{-3x}}{(-3)^2 + 2^2} [(-3) \sin 2x - (2) \cos 2x] + C \)
\( = \frac{e^{-3x}}{9 + 4} [-3 \sin 2x - 2 \cos 2x] + C \)
\( = \frac{e^{-3x}}{13} [-3 \sin 2x - 2 \cos 2x] + C \)
(ii) To integrate \( e^{-4x} \sin 2x \), we use the standard integral formula for \( \int e^{ax} \sin bx \, dx \). This formula is \( \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \). Here, \( a \) is -4 and \( b \) is 2. By substituting these values into the formula and simplifying, we get the required integral. Factoring out common terms from the bracket helps in presenting a more simplified final expression.
Here \( a = -4, b = 2 \)
\( \int e^{-4x} \sin 2x \, dx = \frac{e^{-4x}}{(-4)^2 + 2^2} [(-4) \sin 2x - (2) \cos 2x] + C \)
\( = \frac{e^{-4x}}{16 + 4} [-4 \sin 2x - 2 \cos 2x] + C \)
\( = \frac{e^{-4x}}{20} [-2 (2 \sin 2x + \cos 2x)] + C \)
\( = -\frac{e^{-4x}}{10} [2 \sin 2x + \cos 2x] + C \)
(iii) To integrate \( e^{-3x} \cos x \), we apply the general formula for \( \int e^{ax} \cos bx \, dx \). The formula is \( \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C \). For this problem, \( a \) is -3 and \( b \) is 1. We substitute these values into the formula and simplify the expression to obtain the final integral, making sure to arrange terms clearly.
Here \( a = -3, b = 1 \)
\( \int e^{-3x} \cos x \, dx = \frac{e^{-3x}}{(-3)^2 + 1^2} [(-3) \cos x + (1) \sin x] + C \)
\( = \frac{e^{-3x}}{9 + 1} [\sin x - 3 \cos x] + C \)
\( = \frac{e^{-3x}}{10} [\sin x - 3 \cos x] + C \)
In simple words: These problems involve integrating special combinations of 'e to the power of x' and 'sin x' or 'cos x'. We use a direct mathematical formula by finding the 'a' and 'b' numbers from the question and putting them into the formula to solve.

🎯 Exam Tip: Always double-check the values of 'a' and 'b' and their signs when substituting into the integral formulas, as a small error can lead to a completely different answer.