GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.723

Integrate the following functions:

Question 1. \( \sqrt{4-x^{2}} \)
Answer: We need to integrate the function \( \sqrt{4-x^{2}} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{4-x^{2}} dx \)
We use the formula: \( \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C \)
Here, \( a^{2} = 4 \implies a = 2 \).
So, \( \int \sqrt{4-x^{2}} dx = \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \frac{x}{2} + C \)
\( = \frac{x}{2} \sqrt{4-x^{2}} + 2 \sin^{-1} \frac{x}{2} + C \)
In simple words: To find the integral of `\(\sqrt{4-x^{2}}\)`, we use a specific math rule for square root expressions. We put the values into the formula and simplify to get the final answer, which includes a sine inverse term.

Exam Tip: Remember the standard integration formulas for expressions involving square roots. Identify 'a' correctly, then substitute its value into the formula and simplify for accuracy.

Question 2. \( \sqrt{1-4x^{2}} \)
Answer: We need to integrate the function \( \sqrt{1-4x^{2}} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{1-4x^{2}} dx \)
\( = \int \sqrt{1-(2x)^{2}} dx \)
Let \( 2x = t \).
\( \implies 2 dx = dt \)
\( \implies dx = \frac{1}{2} dt \)
So, \( \int \sqrt{1-4x^{2}} dx = \int \sqrt{1-t^{2}} \frac{1}{2} dt = \frac{1}{2} \int \sqrt{1-t^{2}} dt \)
Using the formula \( \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C \), with \( a=1 \) and \( t \) in place of \( x \):
\( = \frac{1}{2} \left[ \frac{t}{2} \sqrt{1-t^{2}} + \frac{1^{2}}{2} \sin^{-1} \frac{t}{1} \right] + C \)
\( = \frac{1}{2} \left[ \frac{2x}{2} \sqrt{1-(2x)^{2}} + \frac{1}{2} \sin^{-1} (2x) \right] + C \)
\( = \frac{1}{2} \left[ x \sqrt{1-4x^{2}} + \frac{1}{2} \sin^{-1} (2x) \right] + C \)
\( = \frac{x}{2} \sqrt{1-4x^{2}} + \frac{1}{4} \sin^{-1} (2x) + C \)
In simple words: To integrate this expression, we first change `\(4x^{2}\)` into `\((2x)^{2}\)`. Then, we replace `\(2x\)` with a new variable and solve the integral using a standard formula for square root functions. Finally, we put the original `\(x\)` back to get the answer.

Exam Tip: For integrals with `\(ax^2\)` inside the square root, remember to use substitution, like `\(u = \sqrt{a}x\)`, before applying the standard integral formulas. This simplifies the expression and prevents mistakes.

Question 3. \( \sqrt{x^{2}+4x+6} \)
Answer: We need to integrate the function \( \sqrt{x^{2}+4x+6} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{x^{2}+4x+6} dx \)
First, we complete the square inside the square root:
\( x^{2}+4x+6 = x^{2}+4x+4+2 = (x+2)^{2} + (\sqrt{2})^{2} \)
So, \( \int \sqrt{x^{2}+4x+6} dx = \int \sqrt{(x+2)^{2} + (\sqrt{2})^{2}} dx \)
Using the formula \( \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \log|x + \sqrt{x^{2}+a^{2}}| + C \), with \( x \) replaced by \( (x+2) \) and \( a = \sqrt{2} \):
\( = \frac{(x+2)}{2} \sqrt{(x+2)^{2} + (\sqrt{2})^{2}} + \frac{(\sqrt{2})^{2}}{2} \log|(x+2) + \sqrt{(x+2)^{2} + (\sqrt{2})^{2}}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x+4+2} + \frac{2}{2} \log|(x+2) + \sqrt{x^{2}+4x+4+2}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x+6} + \log|(x+2) + \sqrt{x^{2}+4x+6}| + C \)
In simple words: To integrate this, we first change the expression inside the square root by completing the square. This helps us use a standard integration formula for `\(\sqrt{x^{2}+a^{2}}\)`. Then, we just substitute the values back and simplify.

Exam Tip: When you see a quadratic expression under a square root, your first step should always be completing the square to bring it into one of the standard integral forms.

Question 4. \( \sqrt{x^{2}+4x+1} \)
Answer: We need to integrate the function \( \sqrt{x^{2}+4x+1} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{x^{2}+4x+1} dx \)
First, we complete the square inside the square root:
\( x^{2}+4x+1 = x^{2}+4x+4-3 = (x+2)^{2} - (\sqrt{3})^{2} \)
So, \( \int \sqrt{x^{2}+4x+1} dx = \int \sqrt{(x+2)^{2} - (\sqrt{3})^{2}} dx \)
Using the formula \( \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log|x + \sqrt{x^{2}-a^{2}}| + C \), with \( x \) replaced by \( (x+2) \) and \( a = \sqrt{3} \):
\( = \frac{(x+2)}{2} \sqrt{(x+2)^{2} - (\sqrt{3})^{2}} - \frac{(\sqrt{3})^{2}}{2} \log|(x+2) + \sqrt{(x+2)^{2} - (\sqrt{3})^{2}}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x+4-3} - \frac{3}{2} \log|(x+2) + \sqrt{x^{2}+4x+4-3}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x+1} - \frac{3}{2} \log|(x+2) + \sqrt{x^{2}+4x+1}| + C \)
In simple words: First, we change the expression under the square root by completing the square to make it fit a standard formula. Then, we use the formula for `\(\sqrt{x^{2}-a^{2}}\)` by replacing `\(x\)` with `\((x+2)\)` and `\(a\)` with `\(\sqrt{3}\)`. Finally, we simplify the result.

Exam Tip: Pay close attention to the sign before \(a^2\) after completing the square, as it determines which standard formula `\(\int \sqrt{x^2+a^2} dx\)` or `\(\int \sqrt{x^2-a^2} dx\)` to use.

Question 5. \( \sqrt{1-4 x-x^{2}} \)
Answer: We need to integrate the function \( \sqrt{1-4x-x^{2}} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{1-4x-x^{2}} dx \)
First, we complete the square inside the square root:
\( 1-4x-x^{2} = 1-(x^{2}+4x) = 1-(x^{2}+4x+4-4) = 1-( (x+2)^{2} - 4) = 1-(x+2)^{2} + 4 = 5-(x+2)^{2} = (\sqrt{5})^{2}-(x+2)^{2} \)
So, \( \int \sqrt{1-4x-x^{2}} dx = \int \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} dx \)
Using the formula \( \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C \), with \( x \) replaced by \( (x+2) \) and \( a = \sqrt{5} \):
\( = \frac{(x+2)}{2} \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} + \frac{(\sqrt{5})^{2}}{2} \sin^{-1} \frac{(x+2)}{\sqrt{5}} + C \)
\( = \frac{(x+2)}{2} \sqrt{5-(x+2)^{2}} + \frac{5}{2} \sin^{-1} \frac{(x+2)}{\sqrt{5}} + C \)
\( = \frac{(x+2)}{2} \sqrt{1-4x-x^{2}} + \frac{5}{2} \sin^{-1} \frac{(x+2)}{\sqrt{5}} + C \)
In simple words: To solve this integral, we first rearrange the expression under the square root by taking out a minus sign and completing the square for the `\(x\)` terms. This helps us fit it into the standard `\(\sqrt{a^{2}-x^{2}}\)` integral formula, after which we substitute and simplify.

Exam Tip: Be careful with the negative sign when completing the square for expressions like `\(-x^2 - 4x + 1\)`. Factor out the negative sign first, then complete the square for the positive quadratic part.

Question 6. \( \sqrt{x^{2}+4 x-5} \)
Answer: We need to integrate the function \( \sqrt{x^{2}+4x-5} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{x^{2}+4x-5} dx \)
First, we complete the square inside the square root:
\( x^{2}+4x-5 = x^{2}+4x+4-9 = (x+2)^{2} - (3)^{2} \)
So, \( \int \sqrt{x^{2}+4x-5} dx = \int \sqrt{(x+2)^{2} - (3)^{2}} dx \)
Using the formula \( \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log|x + \sqrt{x^{2}-a^{2}}| + C \), with \( x \) replaced by \( (x+2) \) and \( a = 3 \):
\( = \frac{(x+2)}{2} \sqrt{(x+2)^{2} - (3)^{2}} - \frac{3^{2}}{2} \log|(x+2) + \sqrt{(x+2)^{2} - (3)^{2}}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x+4-9} - \frac{9}{2} \log|(x+2) + \sqrt{x^{2}+4x+4-9}| + C \)
\( = \frac{(x+2)}{2} \sqrt{x^{2}+4x-5} - \frac{9}{2} \log|(x+2) + \sqrt{x^{2}+4x-5}| + C \)
In simple words: First, we modify the expression under the square root by completing the square to get it into a simpler form. Then, we use the specific integration formula for `\(\sqrt{x^{2}-a^{2}}\)` and substitute the `\(x\)` and `\(a\)` values back into the formula, then simplify the result.

Exam Tip: Always double-check your arithmetic when completing the square, especially with the constant term. A small error there can lead to using the wrong formula or incorrect final values.

Question 7. \( \sqrt{1+3 x-x^{2}} \)
Answer: We need to integrate the function \( \sqrt{1+3x-x^{2}} \) with respect to \( x \). The solution is presented as follows:
\( \int \sqrt{1+3x-x^{2}} dx \)
First, we complete the square inside the square root:
\( 1+3x-x^{2} = 1-(x^{2}-3x) = 1-(x^{2}-3x+\frac{9}{4}-\frac{9}{4}) = 1-((x-\frac{3}{2})^{2} - \frac{9}{4}) = 1-(x-\frac{3}{2})^{2} + \frac{9}{4} \)
\( = \frac{4+9}{4} - (x-\frac{3}{2})^{2} = \frac{13}{4} - (x-\frac{3}{2})^{2} = (\frac{\sqrt{13}}{2})^{2} - (x-\frac{3}{2})^{2} \)
So, \( \int \sqrt{1+3x-x^{2}} dx = \int \sqrt{(\frac{\sqrt{13}}{2})^{2} - (x-\frac{3}{2})^{2}} dx \)
Using the formula \( \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C \), with \( x \) replaced by \( (x-\frac{3}{2}) \) and \( a = \frac{\sqrt{13}}{2} \):
\( = \frac{(x-\frac{3}{2})}{2} \sqrt{(\frac{\sqrt{13}}{2})^{2} - (x-\frac{3}{2})^{2}} + \frac{(\frac{\sqrt{13}}{2})^{2}}{2} \sin^{-1} \frac{(x-\frac{3}{2})}{(\frac{\sqrt{13}}{2})} + C \)
\( = \frac{(2x-3)}{4} \sqrt{1+3x-x^{2}} + \frac{\frac{13}{4}}{2} \sin^{-1} \frac{(2x-3)}{\sqrt{13}} + C \)
\( = \frac{(2x-3)}{4} \sqrt{1+3x-x^{2}} + \frac{13}{8} \sin^{-1} \frac{(2x-3)}{\sqrt{13}} + C \)
In simple words: To integrate this, we first rework the expression under the square root by completing the square and combining numbers. This transformation allows us to use the standard integral formula for `\(\sqrt{a^{2}-x^{2}}\)`. We then substitute the values into the formula and simplify to get the final answer.

Exam Tip: Handling fractions when completing the square requires careful calculation. Always ensure your common denominators are correct to avoid errors in the `\(a^2\)` term.

Question 8. \( \sqrt{x^{2}+3 x} \)
Answer: We need to integrate the function \( \sqrt{x^{2}+3x} \) with respect to \( x \). The solution is presented as follows:
Let \( I = \int \sqrt{x^{2}+3x} dx \)
First, we complete the square inside the square root:
\( x^{2}+3x = x^{2}+3x+\frac{9}{4}-\frac{9}{4} = (x+\frac{3}{2})^{2} - (\frac{3}{2})^{2} \)
So, \( I = \int \sqrt{(x+\frac{3}{2})^{2} - (\frac{3}{2})^{2}} dx \)
Using the formula \( \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log|x + \sqrt{x^{2}-a^{2}}| + C \), with \( x \) replaced by \( (x+\frac{3}{2}) \) and \( a = \frac{3}{2} \):
\( = \frac{(x+\frac{3}{2})}{2} \sqrt{(x+\frac{3}{2})^{2} - (\frac{3}{2})^{2}} - \frac{(\frac{3}{2})^{2}}{2} \log|(x+\frac{3}{2}) + \sqrt{(x+\frac{3}{2})^{2} - (\frac{3}{2})^{2}}| + C \)
\( = \frac{(2x+3)}{4} \sqrt{x^{2}+3x} - \frac{\frac{9}{4}}{2} \log|(x+\frac{3}{2}) + \sqrt{x^{2}+3x}| + C \)
\( = \frac{(2x+3)}{4} \sqrt{x^{2}+3x} - \frac{9}{8} \log|(x+\frac{3}{2}) + \sqrt{x^{2}+3x}| + C \)
In simple words: We start by rewriting the expression inside the square root by completing the square. This lets us use the standard integration formula for `\(\sqrt{x^{2}-a^{2}}\)`. After applying the formula and substituting values, we simplify to reach the final answer.

Exam Tip: Always make sure to replace all instances of 'x' in the formula with the transformed expression (e.g., `\(x+\frac{3}{2}\)`), not just the first one. This is a common point of error.

Question 9. \( \sqrt{1+\frac{x^{2}}{9}} \)
Answer: We need to integrate the function \( \sqrt{1+\frac{x^{2}}{9}} \) with respect to \( x \). The solution is presented as follows:
Let \( I = \int \sqrt{1+\frac{x^{2}}{9}} dx \)
We can rewrite the expression as: \( \int \sqrt{\frac{9+x^{2}}{9}} dx = \frac{1}{3} \int \sqrt{x^{2}+9} dx \)
This is in the form \( \int \sqrt{x^{2} + a^{2}} dx \). Here, \( a^{2} = 9 \implies a = 3 \).
Using the formula \( \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \log|x + \sqrt{x^{2}+a^{2}}| + C \):
\( I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{x^{2}+9} + \frac{9}{2} \log|x + \sqrt{x^{2}+9}| \right] + C \)
\( = \frac{1}{6} \left[ x \sqrt{x^{2}+9} + 9 \log|x + \sqrt{x^{2}+9}| \right] + C \)
In simple words: First, we combine the terms under the square root and bring the constant out. Then, we identify the value of `\(a\)` and use the standard integration formula for `\(\sqrt{x^{2}+a^{2}}\)`. Finally, we multiply by the outside constant to get the full answer.

Exam Tip: Always simplify the expression under the square root first, if possible, to reveal the correct form (e.g., `\(\sqrt{a^2 \pm x^2}\)` or `\(\sqrt{x^2 \pm a^2}\)`) for applying the standard formulas.

Choose the correct answers in questions 10 and 11:

Question 10. \( \int \sqrt{1+x^{2}} dx \) is equal to
(a) \( \frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \log|x + \sqrt{1+x^{2}}| + C \)
(b) \( \frac{2}{3}(1 + x^{2})^{3/2} + C \)
(c) \( \frac{2}{3}x(1 + x^{2})^{3/2} + C \)
(d) \( \frac{x^{2}}{2}\sqrt{1+x^{2}} + \frac{1}{2} \log|x + \sqrt{1+x^{2}}| + C \)
Answer: (a) \( \frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \log|x + \sqrt{1+x^{2}}| + C \)
In simple words: The integral of `\(\sqrt{1+x^{2}}\)` is a standard formula. We apply the formula `\(\int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \log|x + \sqrt{x^{2}+a^{2}}| + C\)` with `\(a=1\)` to get the correct option.

Exam Tip: Recognize standard integral forms like `\(\int \sqrt{a^2+x^2} dx\)` and recall their formulas directly. This question directly tests knowledge of these standard formulas.

Question 11. \( \int \sqrt{x^{2}-8x+7} dx \) is equal to
(a) \( \frac{1}{2}(x - 4)\sqrt{x^{2}-8 x+7} + 9\log |x - 4 + \sqrt{x^{2}-8 x+7}| + C \)
(b) \( \frac{1}{2}(x + 4)\sqrt{x^{2}-8 x+7} + 9\log |x + 4 + \sqrt{x^{2}-8 x+7}| + C \)
(c) \( \frac{1}{2}(x - 4)\sqrt{x^{2}-8 x+7} – 3\sqrt{2}\log |x - 4 + \sqrt{x^{2}-8 x+7}| + C \)
(d) \( \frac{1}{2}(x - 4)\sqrt{x^{2}-8 x+7} – \frac{9}{2}\log |x - 4 + \sqrt{x^{2}-8 x+7}| + C \)
Answer: (d) \( \frac{1}{2}(x - 4)\sqrt{x^{2}-8 x+7} – \frac{9}{2}\log |x - 4 + \sqrt{x^{2}-8 x+7}| + C \)
In simple words: We first complete the square for `\(x^{2}-8x+7\)` to get `\((x-4)^{2}-9\)` or `\((x-4)^{2}-3^{2}\)`. Then, we apply the integration formula for `\(\sqrt{x^{2}-a^{2}}\)` by replacing `\(x\)` with `\((x-4)\)` and `\(a\)` with `\(3\)` to match the correct option.

Exam Tip: For MCQ questions involving integrals of quadratic expressions under square roots, complete the square first. This will help you identify the correct standard form and select the right option by comparing the terms and signs.