OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (F)

Question 1. \( \int \sqrt{1-x^2} d x \)
Answer: To solve this integral, we use the standard formula for \( \int \sqrt{a^2-x^2} d x \). In this case, \( a^2 = 1 \), so \( a = 1 \).
The formula is: \( \int \sqrt{a^2-x^2} d x = \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \)
Substituting \( a=1 \) into the formula:
\( \int \sqrt{1^2-x^2} d x = \frac{x \sqrt{1-x^2}}{2} + \frac{1^2}{2}\sin^{-1}\left(\frac{x}{1}\right) + C \)
\( \implies \int \sqrt{1-x^2} d x = \frac{x \sqrt{1-x^2}}{2} + \frac{1}{2}\sin^{-1}(x) + C \)
In simple words: This integral is a standard type. We just need to put the value of 'a' into the ready-made formula to get the answer.

🎯 Exam Tip: Memorize the three main formulas for integrals involving square roots: \( \int \sqrt{a^2-x^2} dx \), \( \int \sqrt{x^2+a^2} dx \), and \( \int \sqrt{x^2-a^2} dx \) to save time in exams.

Question 2. \( \int \sqrt{4-x^2} d x \)
Answer: We need to integrate \( \sqrt{4-x^2} \). We can write 4 as \( 2^2 \).
So, the integral becomes \( \int \sqrt{2^2-x^2} d x \).
This matches the standard formula \( \int \sqrt{a^2-x^2} d x = \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \), where \( a=2 \).
Substituting \( a=2 \) into the formula:
\( \int \sqrt{2^2-x^2} d x = \frac{x \sqrt{2^2-x^2}}{2} + \frac{2^2}{2}\sin^{-1}\left(\frac{x}{2}\right) + C \)
\( \implies \int \sqrt{4-x^2} d x = \frac{x \sqrt{4-x^2}}{2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) + C \)
\( \implies \int \sqrt{4-x^2} d x = \frac{x \sqrt{4-x^2}}{2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C \)
In simple words: First, rewrite 4 as 2 squared. Then, use the standard integral formula for square roots, replacing 'a' with 2.

🎯 Exam Tip: Always identify the value of 'a' correctly from the constant term in the integral before applying the formula. Simplifying fractions like \( a^2/2 \) is also important.

Question 3. \( \int \sqrt{3-4 x^2} d x \)
Answer: To integrate \( \sqrt{3-4 x^2} \), we first need to make the coefficient of \( x^2 \) equal to 1. We do this by factoring out 4 from the term under the square root.
\( \int \sqrt{3-4 x^2} d x = \int \sqrt{4\left(\frac{3}{4}-x^2\right)} d x \)
\( \implies = \sqrt{4} \int \sqrt{\frac{3}{4}-x^2} d x \)
\( \implies = 2 \int \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2-x^2} d x \)
Now, this is in the form \( \int \sqrt{a^2-x^2} d x \), where \( a = \frac{\sqrt{3}}{2} \).
Using the formula \( \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \):
\( = 2 \left[ \frac{x \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2-x^2}}{2} + \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2}\sin^{-1}\left(\frac{x}{\frac{\sqrt{3}}{2}}\right) \right] + C \)
\( = 2 \left[ \frac{x \sqrt{\frac{3}{4}-x^2}}{2} + \frac{\frac{3}{4}}{2}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) \right] + C \)
\( = 2 \left[ \frac{x \sqrt{\frac{3-4x^2}{4}}}{2} + \frac{3}{8}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) \right] + C \)
\( = 2 \left[ \frac{x \frac{\sqrt{3-4x^2}}{2}}{2} + \frac{3}{8}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) \right] + C \)
\( = 2 \left[ \frac{x \sqrt{3-4x^2}}{4} + \frac{3}{8}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) \right] + C \)
\( \implies = \frac{x \sqrt{3-4x^2}}{2} + \frac{3}{4}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) + C \)
In simple words: First, take out 4 from under the square root so that \( x^2 \) is alone. Then, the number left will be 'a' squared. Use the standard formula to find the answer.

🎯 Exam Tip: When the \( x^2 \) term has a coefficient, always factor it out from under the square root before applying any standard integral formula. This often involves adjusting the constant term as well.

Question 4. \( \int \sqrt{1+x^2} d x \)
Answer: This integral is of the form \( \int \sqrt{a^2+x^2} d x \). Here, \( a^2=1 \), so \( a=1 \).
The standard formula for this type of integral is:
\( \int \sqrt{a^2+x^2} d x = \frac{x \sqrt{a^2+x^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{a^2+x^2}| + C \)
Substituting \( a=1 \) into the formula:
\( \int \sqrt{1^2+x^2} d x = \frac{x \sqrt{1+x^2}}{2} + \frac{1^2}{2}\log|x + \sqrt{1+x^2}| + C \)
\( \implies \int \sqrt{1+x^2} d x = \frac{x \sqrt{1+x^2}}{2} + \frac{1}{2}\log|x + \sqrt{1+x^2}| + C \)
In simple words: This is another common integral formula. Just identify that 'a' is 1, and then plug it into the correct formula for square roots with a plus sign.

🎯 Exam Tip: Differentiate between \( \int \sqrt{a^2-x^2} dx \) (involving \( \sin^{-1} \)) and \( \int \sqrt{x^2+a^2} dx \) or \( \int \sqrt{x^2-a^2} dx \) (involving \( \log \) functions) to apply the correct formula.

Question 5. \( \int \sqrt{16+x^2} d x \)
Answer: The given integral is \( \int \sqrt{16+x^2} d x \). We can rewrite 16 as \( 4^2 \).
So, the integral becomes \( \int \sqrt{4^2+x^2} d x \).
This is in the standard form \( \int \sqrt{a^2+x^2} d x \), where \( a=4 \).
Using the formula: \( \int \sqrt{a^2+x^2} d x = \frac{x \sqrt{a^2+x^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{a^2+x^2}| + C \)
Substituting \( a=4 \) into the formula:
\( \int \sqrt{4^2+x^2} d x = \frac{x \sqrt{4^2+x^2}}{2} + \frac{4^2}{2}\log|x + \sqrt{4^2+x^2}| + C \)
\( \implies = \frac{x \sqrt{16+x^2}}{2} + \frac{16}{2}\log|x + \sqrt{16+x^2}| + C \)
\( \implies = \frac{x \sqrt{16+x^2}}{2} + 8\log|x + \sqrt{16+x^2}| + C \)
In simple words: First, change 16 into 4 squared. Then, use the general integral formula for \( a^2+x^2 \) by putting 4 in place of 'a'.

🎯 Exam Tip: Always look for perfect squares (like 16, 25, 36) to identify 'a' quickly. If the constant is not a perfect square, express it as \( (\sqrt{k})^2 \).

Question 6. \( \int \sqrt{x^2-36} d x \)
Answer: The integral is \( \int \sqrt{x^2-36} d x \). We can write 36 as \( 6^2 \).
So, the integral becomes \( \int \sqrt{x^2-6^2} d x \).
This is in the standard form \( \int \sqrt{x^2-a^2} d x \), where \( a=6 \).
The formula for this integral is:
\( \int \sqrt{x^2-a^2} d x = \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \)
Substituting \( a=6 \) into the formula:
\( \int \sqrt{x^2-6^2} d x = \frac{x \sqrt{x^2-6^2}}{2} - \frac{6^2}{2}\log|x + \sqrt{x^2-6^2}| + C \)
\( \implies = \frac{x \sqrt{x^2-36}}{2} - \frac{36}{2}\log|x + \sqrt{x^2-36}| + C \)
\( \implies = \frac{x \sqrt{x^2-36}}{2} - 18\log|x + \sqrt{x^2-36}| + C \)
In simple words: Change 36 into 6 squared. Then, use the special formula for integrals with \( x^2 \) minus a number, where 'a' is 6. Remember the minus sign in the log term.

🎯 Exam Tip: Pay close attention to the sign between \( x^2 \) and \( a^2 \). A minus sign before \( a^2 \) leads to a different formula (and a minus sign in the result's logarithmic term) than a plus sign.

Question 7. \( \int \sqrt{3 x^2+5} d x \)
Answer: To integrate \( \sqrt{3 x^2+5} \), we first need to make the coefficient of \( x^2 \) equal to 1. We do this by factoring out \( \sqrt{3} \) from the whole expression.
\( \int \sqrt{3 x^2+5} d x = \int \sqrt{3\left(x^2+\frac{5}{3}\right)} d x \)
\( \implies = \sqrt{3} \int \sqrt{x^2+\frac{5}{3}} d x \)
We can write \( \frac{5}{3} \) as \( \left(\sqrt{\frac{5}{3}}\right)^2 \).
So, the integral becomes \( \sqrt{3} \int \sqrt{x^2+\left(\sqrt{\frac{5}{3}}\right)^2} d x \).
This is in the form \( \int \sqrt{x^2+a^2} d x \), where \( a = \sqrt{\frac{5}{3}} \).
Using the formula \( \int \sqrt{x^2+a^2} d x = \frac{x \sqrt{x^2+a^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{x^2+a^2}| + C \):
\( = \sqrt{3} \left[ \frac{x \sqrt{x^2+\frac{5}{3}}}{2} + \frac{\frac{5}{3}}{2}\log\left|x + \sqrt{x^2+\frac{5}{3}}\right| \right] + C \)
\( = \sqrt{3} \left[ \frac{x \sqrt{x^2+\frac{5}{3}}}{2} + \frac{5}{6}\log\left|x + \sqrt{x^2+\frac{5}{3}}\right| \right] + C \)
Now, we multiply \( \sqrt{3} \) inside the bracket:
\( = \frac{x \sqrt{3}\sqrt{x^2+\frac{5}{3}}}{2} + \frac{5\sqrt{3}}{6}\log\left|x + \sqrt{x^2+\frac{5}{3}}\right| + C \)
\( = \frac{x \sqrt{3x^2+5}}{2} + \frac{5\sqrt{3}}{6}\log\left|x + \sqrt{x^2+\frac{5}{3}}\right| + C \)
In simple words: First, pull out \( \sqrt{3} \) from the whole integral so that \( x^2 \) is alone. Then, treat \( 5/3 \) as 'a' squared and use the standard formula for \( x^2+a^2 \). Remember to multiply the \( \sqrt{3} \) back at the end.

🎯 Exam Tip: When a constant multiplies \( x^2 \) under the square root, always factor it out. This often means the 'a' value will be a fraction or involve a square root, requiring careful calculation.

Question 8. \( \int x \sqrt{x^4+1} d x \)
Answer: To solve this integral, we use substitution. Let \( t = x^2 \).
Then, differentiate \( t \) with respect to \( x \): \( dt = 2x \, dx \).
This means \( x \, dx = \frac{1}{2} dt \).
Also, \( x^4 = (x^2)^2 = t^2 \).
Substituting these into the integral:
\( I = \int \sqrt{t^2+1} \cdot \frac{1}{2} dt \)
\( \implies I = \frac{1}{2} \int \sqrt{t^2+1^2} dt \)
This is in the standard form \( \int \sqrt{x^2+a^2} d x \), where \( x=t \) and \( a=1 \).
Using the formula \( \frac{x \sqrt{x^2+a^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{x^2+a^2}| + C \):
\( I = \frac{1}{2} \left[ \frac{t \sqrt{t^2+1}}{2} + \frac{1^2}{2}\log|t + \sqrt{t^2+1}| \right] + C \)
\( \implies I = \frac{1}{2} \left[ \frac{t \sqrt{t^2+1}}{2} + \frac{1}{2}\log|t + \sqrt{t^2+1}| \right] + C \)
\( \implies I = \frac{t \sqrt{t^2+1}}{4} + \frac{1}{4}\log|t + \sqrt{t^2+1}| + C \)
Now, substitute back \( t = x^2 \):
\( \implies I = \frac{x^2 \sqrt{(x^2)^2+1}}{4} + \frac{1}{4}\log|x^2 + \sqrt{(x^2)^2+1}| + C \)
\( \implies I = \frac{x^2 \sqrt{x^4+1}}{4} + \frac{1}{4}\log|x^2 + \sqrt{x^4+1}| + C \)
In simple words: Use substitution by letting \( x^2 \) be 't'. This makes the integral look like a standard formula. Solve it, then put \( x^2 \) back where 't' was.

🎯 Exam Tip: When you see \( x^4 \) or \( x^2 \) along with an \( x \) outside the square root, consider substitution using \( t = x^2 \) as it often simplifies the integral into a standard form.

Question 9. \( \int \frac{\sqrt{1+(\log x)^2}}{x} d x \)
Answer: We will use substitution to solve this integral. Let \( t = \log x \).
Now, differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{x} dx \).
Substitute these into the integral:
\( I = \int \sqrt{1+t^2} dt \)
This is in the standard form \( \int \sqrt{a^2+x^2} d x \), where \( x=t \) and \( a=1 \).
Using the formula \( \int \sqrt{a^2+x^2} d x = \frac{x \sqrt{a^2+x^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{a^2+x^2}| + C \):
\( I = \frac{t \sqrt{t^2+1}}{2} + \frac{1^2}{2}\log|t + \sqrt{t^2+1}| + C \)
\( \implies I = \frac{t \sqrt{t^2+1}}{2} + \frac{1}{2}\log|t + \sqrt{t^2+1}| + C \)
Now, substitute back \( t = \log x \):
\( \implies I = \frac{\log x \sqrt{(\log x)^2+1}}{2} + \frac{1}{2}\log|\log x + \sqrt{(\log x)^2+1}| + C \)
In simple words: Let \( \log x \) be 't'. Then, \( 1/x \, dx \) becomes 'dt'. This simplifies the problem to a standard integral form. After solving, replace 't' with \( \log x \) everywhere.

🎯 Exam Tip: Always look for a function and its derivative within the integral. If a derivative (like \( 1/x \) for \( \log x \)) is present, substitution is usually the most effective method.

Question 10. \( \int \sqrt{x^2+4x+6} d x \)
Answer: To solve this integral, we first need to complete the square for the expression inside the square root: \( x^2+4x+6 \).
\( x^2+4x+6 = x^2+4x+4+2 \)
\( \implies = (x+2)^2+2 \)
We can write 2 as \( (\sqrt{2})^2 \).
So, the expression becomes \( (x+2)^2+(\sqrt{2})^2 \).
Now, the integral is \( \int \sqrt{(x+2)^2+(\sqrt{2})^2} d x \).
Let \( t = x+2 \). Then, differentiate \( t \) with respect to \( x \): \( dt = dx \).
The integral becomes \( \int \sqrt{t^2+(\sqrt{2})^2} dt \).
This is in the standard form \( \int \sqrt{x^2+a^2} d x \), where \( x=t \) and \( a=\sqrt{2} \).
Using the formula \( \frac{x \sqrt{x^2+a^2}}{2} + \frac{a^2}{2}\log|x + \sqrt{x^2+a^2}| + C \):
\( = \frac{t \sqrt{t^2+(\sqrt{2})^2}}{2} + \frac{(\sqrt{2})^2}{2}\log|t + \sqrt{t^2+(\sqrt{2})^2}| + C \)
\( = \frac{t \sqrt{t^2+2}}{2} + \frac{2}{2}\log|t + \sqrt{t^2+2}| + C \)
\( \implies = \frac{t \sqrt{t^2+2}}{2} + \log|t + \sqrt{t^2+2}| + C \)
Finally, substitute back \( t=x+2 \):
\( \implies = \frac{(x+2) \sqrt{(x+2)^2+2}}{2} + \log|x+2+\sqrt{(x+2)^2+2}| + C \)
\( \implies = \frac{(x+2) \sqrt{x^2+4x+6}}{2} + \log|x+2+\sqrt{x^2+4x+6}| + C \)
In simple words: First, change the part under the square root by completing the square to get \( (x+k)^2+a^2 \). Then, use substitution with \( t=x+k \) and solve with the standard formula. Remember to put \( x+k \) back in for 't' at the end.

🎯 Exam Tip: Completing the square is crucial for integrals involving quadratic expressions under the square root. Always verify your completed square form to match one of the standard integral types.

Question 11. \( \int \sqrt{x^2+3x} d x \)
Answer: We need to complete the square for the expression inside the square root: \( x^2+3x \).
To complete the square for \( x^2+3x \), we add and subtract \( \left(\frac{3}{2}\right)^2 \):
\( x^2+3x = x^2+3x+\left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 \)
\( \implies = \left(x+\frac{3}{2}\right)^2 - \frac{9}{4} \)
So, the integral becomes \( \int \sqrt{\left(x+\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2} d x \).
Let \( t = x+\frac{3}{2} \). Then, \( dt = dx \).
The integral becomes \( \int \sqrt{t^2-\left(\frac{3}{2}\right)^2} dt \).
This is in the standard form \( \int \sqrt{x^2-a^2} d x \), where \( x=t \) and \( a=\frac{3}{2} \).
Using the formula \( \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( = \frac{t \sqrt{t^2-\left(\frac{3}{2}\right)^2}}{2} - \frac{\left(\frac{3}{2}\right)^2}{2}\log\left|t + \sqrt{t^2-\left(\frac{3}{2}\right)^2}\right| + C \)
\( = \frac{t \sqrt{t^2-\frac{9}{4}}}{2} - \frac{\frac{9}{4}}{2}\log\left|t + \sqrt{t^2-\frac{9}{4}}\right| + C \)
\( \implies = \frac{t \sqrt{t^2-\frac{9}{4}}}{2} - \frac{9}{8}\log\left|t + \sqrt{t^2-\frac{9}{4}}\right| + C \)
Now, substitute back \( t = x+\frac{3}{2} \):
\( = \frac{\left(x+\frac{3}{2}\right) \sqrt{\left(x+\frac{3}{2}\right)^2-\frac{9}{4}}}{2} - \frac{9}{8}\log\left|x+\frac{3}{2}+\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{9}{4}}\right| + C \)
\( \implies = \frac{\frac{2x+3}{2} \sqrt{x^2+3x}}{2} - \frac{9}{8}\log\left|x+\frac{3}{2}+\sqrt{x^2+3x}\right| + C \)
\( \implies = \frac{(2x+3) \sqrt{x^2+3x}}{4} - \frac{9}{8}\log\left|x+\frac{3}{2}+\sqrt{x^2+3x}\right| + C \)
In simple words: Complete the square for \( x^2+3x \) to get \( (x+3/2)^2 - (3/2)^2 \). Then, use substitution and apply the standard integral formula for \( t^2-a^2 \). Remember to convert \( t \) back to \( x+3/2 \) in the end.

🎯 Exam Tip: When completing the square, remember to add and subtract the square of half the coefficient of x. Be careful with signs, especially if it leads to an \( x^2-a^2 \) form, which has a minus sign in the log term.

Question 12. \( \int \sqrt{1+2 x-3 x^2} d x \)
Answer: To solve this integral, we first need to complete the square for the expression \( 1+2x-3x^2 \).
Factor out -3 from the \( x \) terms:
\( 1+2x-3x^2 = 1 - 3(x^2 - \frac{2}{3}x) \)
Now, complete the square for \( x^2 - \frac{2}{3}x \) by adding and subtracting \( \left(\frac{1}{3}\right)^2 \):
\( = 1 - 3\left(x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2\right) \)
\( = 1 - 3\left(\left(x-\frac{1}{3}\right)^2 - \frac{1}{9}\right) \)
\( = 1 - 3\left(x-\frac{1}{3}\right)^2 + 3 \cdot \frac{1}{9} \)
\( = 1 - 3\left(x-\frac{1}{3}\right)^2 + \frac{1}{3} \)
\( = \frac{4}{3} - 3\left(x-\frac{1}{3}\right)^2 \)
Now, factor out 3 again to match the standard form:
\( = 3\left(\frac{4}{9} - \left(x-\frac{1}{3}\right)^2\right) \)
So the integral becomes \( \int \sqrt{3\left(\frac{4}{9} - \left(x-\frac{1}{3}\right)^2\right)} d x \)
\( = \sqrt{3} \int \sqrt{\left(\frac{2}{3}\right)^2 - \left(x-\frac{1}{3}\right)^2} d x \)
Let \( t = x-\frac{1}{3} \). Then \( dt = dx \). Also, \( a = \frac{2}{3} \).
The integral becomes \( \sqrt{3} \int \sqrt{a^2-t^2} dt \).
Using the formula \( \int \sqrt{a^2-x^2} d x = \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \):
\( = \sqrt{3} \left[ \frac{t \sqrt{a^2-t^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}\right) \right] + C \)
\( = \sqrt{3} \left[ \frac{\left(x-\frac{1}{3}\right) \sqrt{\frac{4}{9}-\left(x-\frac{1}{3}\right)^2}}{2} + \frac{\left(\frac{2}{3}\right)^2}{2}\sin^{-1}\left(\frac{x-\frac{1}{3}}{\frac{2}{3}}\right) \right] + C \)
\( = \sqrt{3} \left[ \frac{\frac{3x-1}{3} \sqrt{\frac{4}{9}-\left(x-\frac{1}{3}\right)^2}}{2} + \frac{\frac{4}{9}}{2}\sin^{-1}\left(\frac{3x-1}{2}\right) \right] + C \)
\( = \sqrt{3} \left[ \frac{(3x-1) \sqrt{\frac{4-(3x-1)^2}{9}}}{6} + \frac{2}{9}\sin^{-1}\left(\frac{3x-1}{2}\right) \right] + C \)
\( = \sqrt{3} \left[ \frac{(3x-1) \sqrt{4-(3x-1)^2}}{18} + \frac{2}{9}\sin^{-1}\left(\frac{3x-1}{2}\right) \right] + C \)
Note that \( 4-(3x-1)^2 = 4-(9x^2-6x+1) = 3+6x-9x^2 = 3(1+2x-3x^2) \).
So, \( \sqrt{4-(3x-1)^2} = \sqrt{3}\sqrt{1+2x-3x^2} \).
\( = \frac{(3x-1)\sqrt{3}\sqrt{1+2x-3x^2}}{18} + \frac{2\sqrt{3}}{9}\sin^{-1}\left(\frac{3x-1}{2}\right) + C \)
\( = \frac{(3x-1)\sqrt{1+2x-3x^2}}{6} + \frac{2\sqrt{3}}{9}\sin^{-1}\left(\frac{3x-1}{2}\right) + C \)
In simple words: This problem needs careful steps. First, rewrite the expression under the square root by factoring out -3 and completing the square. This will get it into the form \( a^2-t^2 \). Then, use substitution and apply the standard integral formula involving \( \sin^{-1} \).

🎯 Exam Tip: When the coefficient of \( x^2 \) is negative, factor out the negative sign first, then complete the square. Remember to put the factored constant back into the final simplified answer.

Question 13. \( \int \sqrt{1-4 x-x^2} d x \)
Answer: To solve this integral, we complete the square for the expression inside the square root: \( 1-4x-x^2 \).
First, factor out the negative sign from the \( x \) terms:
\( 1-4x-x^2 = 1 - (x^2+4x) \)
Complete the square for \( x^2+4x \) by adding and subtracting \( \left(\frac{4}{2}\right)^2 = 2^2=4 \):
\( = 1 - (x^2+4x+4-4) \)
\( = 1 - ((x+2)^2-4) \)
\( = 1 - (x+2)^2 + 4 \)
\( = 5 - (x+2)^2 \)
We can write 5 as \( (\sqrt{5})^2 \).
So, the expression becomes \( (\sqrt{5})^2-(x+2)^2 \).
The integral is now \( \int \sqrt{(\sqrt{5})^2-(x+2)^2} d x \).
Let \( t = x+2 \). Then \( dt = dx \). Also, \( a = \sqrt{5} \).
The integral becomes \( \int \sqrt{a^2-t^2} dt \).
Using the formula \( \int \sqrt{a^2-x^2} d x = \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \):
\( = \frac{t \sqrt{a^2-t^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}\right) + C \)
\( = \frac{(x+2) \sqrt{(\sqrt{5})^2-(x+2)^2}}{2} + \frac{(\sqrt{5})^2}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \)
\( \implies = \frac{(x+2) \sqrt{5-(x+2)^2}}{2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \)
\( \implies = \frac{(x+2) \sqrt{1-4x-x^2}}{2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \)
In simple words: To integrate this, first complete the square for the part under the root. You need to factor out the minus sign from \( x^2 \) and \( x \). This will turn it into a form like \( a^2-t^2 \). Then, use the standard integral formula that has \( \sin^{-1} \) in it.

🎯 Exam Tip: When completing the square and you have \( -(x^2+Bx) \), remember that \( -(x^2+Bx+C-C) = -(x+B/2)^2 + C \). This reversal of signs is critical for determining the final standard form.

Question 14. \( \int \sqrt{x(1-x)} d x \)
Answer: First, expand the expression inside the square root: \( x(1-x) = x-x^2 \).
So the integral is \( \int \sqrt{x-x^2} d x \).
Now, we complete the square for \( x-x^2 \). Factor out the negative sign:
\( x-x^2 = -(x^2-x) \)
Complete the square for \( x^2-x \) by adding and subtracting \( \left(\frac{1}{2}\right)^2 \):
\( = -\left(x^2-x+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2\right) \)
\( = -\left(\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}\right) \)
\( = -\left(x-\frac{1}{2}\right)^2 + \frac{1}{4} \)
\( = \frac{1}{4} - \left(x-\frac{1}{2}\right)^2 \)
We can write \( \frac{1}{4} \) as \( \left(\frac{1}{2}\right)^2 \).
So, the expression becomes \( \left(\frac{1}{2}\right)^2 - \left(x-\frac{1}{2}\right)^2 \).
The integral is now \( \int \sqrt{\left(\frac{1}{2}\right)^2 - \left(x-\frac{1}{2}\right)^2} d x \).
Let \( t = x-\frac{1}{2} \). Then \( dt = dx \). Also, \( a = \frac{1}{2} \).
The integral becomes \( \int \sqrt{a^2-t^2} dt \).
Using the formula \( \int \sqrt{a^2-x^2} d x = \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \):
\( = \frac{t \sqrt{a^2-t^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}\right) + C \)
\( = \frac{\left(x-\frac{1}{2}\right) \sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}{2} + \frac{\left(\frac{1}{2}\right)^2}{2}\sin^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right) + C \)
\( = \frac{\frac{2x-1}{2} \sqrt{x-x^2}}{2} + \frac{\frac{1}{4}}{2}\sin^{-1}\left(\frac{\frac{2x-1}{2}}{\frac{1}{2}}\right) + C \)
\( \implies = \frac{(2x-1) \sqrt{x-x^2}}{4} + \frac{1}{8}\sin^{-1}(2x-1) + C \)
In simple words: First, multiply out \( x(1-x) \). Then, complete the square by taking out the minus sign from \( x^2 \). This will give you the form \( a^2-t^2 \). After that, substitute and use the standard integral formula that has \( \sin^{-1} \) in it.

🎯 Exam Tip: Always expand products under the square root first. Often, they hide a quadratic expression that needs completing the square to fit a standard integral form.

Question 15. \( \int \sqrt{x^2+4x+1} d x \)
Answer: To solve this integral, we complete the square for the expression inside the square root: \( x^2+4x+1 \).
To complete the square for \( x^2+4x+1 \), we add and subtract \( \left(\frac{4}{2}\right)^2 = 2^2=4 \):
\( x^2+4x+1 = x^2+4x+4-4+1 \)
\( \implies = (x+2)^2-3 \)
We can write 3 as \( (\sqrt{3})^2 \).
So, the expression becomes \( (x+2)^2-(\sqrt{3})^2 \).
Now, the integral is \( \int \sqrt{(x+2)^2-(\sqrt{3})^2} d x \).
Let \( t = x+2 \). Then \( dt = dx \). Also, \( a = \sqrt{3} \).
The integral becomes \( \int \sqrt{t^2-a^2} dt \).
Using the formula \( \int \sqrt{x^2-a^2} d x = \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( = \frac{t \sqrt{t^2-a^2}}{2} - \frac{a^2}{2}\log|t + \sqrt{t^2-a^2}| + C \)
\( = \frac{(x+2) \sqrt{(x+2)^2-(\sqrt{3})^2}}{2} - \frac{(\sqrt{3})^2}{2}\log\left|x+2+\sqrt{(x+2)^2-(\sqrt{3})^2}\right| + C \)
\( \implies = \frac{(x+2) \sqrt{x^2+4x+1}}{2} - \frac{3}{2}\log\left|x+2+\sqrt{x^2+4x+1}\right| + C \)
In simple words: Complete the square for the part under the square root to get it into the form \( (x+k)^2-a^2 \). Then, use substitution with \( t=x+k \) and solve using the standard integral formula involving the log function.

🎯 Exam Tip: After completing the square, carefully observe the resulting form (\( a^2-x^2 \), \( x^2+a^2 \), or \( x^2-a^2 \)) to choose the correct standard integration formula and its corresponding logarithmic or inverse trigonometric part.

Question 16. \( \int(2 x+3) \sqrt{x^2+4 x+3} d x \)
Answer: This integral can be solved by splitting it into two parts. Notice that the derivative of \( x^2+4x+3 \) is \( 2x+4 \). We can adjust \( 2x+3 \) to include \( 2x+4 \).
\( \int(2 x+3) \sqrt{x^2+4 x+3} d x = \int(2 x+4-1) \sqrt{x^2+4 x+3} d x \)
\( \implies = \int(2 x+4) \sqrt{x^2+4 x+3} d x - \int \sqrt{x^2+4 x+3} d x \)
Let's solve the first integral, \( I_1 = \int(2 x+4) \sqrt{x^2+4 x+3} d x \).
Let \( u = x^2+4x+3 \). Then \( du = (2x+4) dx \).
So, \( I_1 = \int \sqrt{u} du = \int u^{1/2} du = \frac{u^{3/2}}{3/2} + C_1 = \frac{2}{3}(x^2+4x+3)^{3/2} + C_1 \).
Now, let's solve the second integral, \( I_2 = \int \sqrt{x^2+4 x+3} d x \).
Complete the square for \( x^2+4x+3 \):
\( x^2+4x+3 = x^2+4x+4-1 = (x+2)^2-1^2 \).
This is in the form \( \int \sqrt{t^2-a^2} dt \), where \( t=x+2 \) and \( a=1 \).
Using the formula \( \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( I_2 = \frac{(x+2) \sqrt{(x+2)^2-1}}{2} - \frac{1^2}{2}\log|x+2+\sqrt{(x+2)^2-1}| + C_2 \)
\( \implies = \frac{(x+2) \sqrt{x^2+4x+3}}{2} - \frac{1}{2}\log|x+2+\sqrt{x^2+4x+3}| + C_2 \).
Combine \( I_1 - I_2 \):
\( \int(2 x+3) \sqrt{x^2+4 x+3} d x = \frac{2}{3}(x^2+4x+3)^{3/2} - \left[ \frac{(x+2) \sqrt{x^2+4x+3}}{2} - \frac{1}{2}\log|x+2+\sqrt{x^2+4x+3}| \right] + C \)
\( \implies = \frac{2}{3}(x^2+4x+3)^{3/2} - \frac{(x+2) \sqrt{x^2+4x+3}}{2} + \frac{1}{2}\log|x+2+\sqrt{x^2+4x+3}| + C \)
In simple words: Break this integral into two parts. For the first part, adjust \( (2x+3) \) to be the derivative of \( x^2+4x+3 \), then use simple substitution. For the second part, complete the square and use a standard integral formula.

🎯 Exam Tip: For integrals of the form \( \int (Px+Q)\sqrt{ax^2+bx+c} dx \), always split \( Px+Q \) into \( A(2ax+b)+B \) to make one part solvable by substitution and the other by completing the square.

Question 17. \( \int(2 x-5) \sqrt{x^2-4 x+3} d x \)
Answer: We will solve this integral by splitting it into two parts. The derivative of \( x^2-4x+3 \) is \( 2x-4 \). We can adjust \( 2x-5 \) to include \( 2x-4 \).
\( \int(2 x-5) \sqrt{x^2-4 x+3} d x = \int(2 x-4-1) \sqrt{x^2-4 x+3} d x \)
\( \implies = \int(2 x-4) \sqrt{x^2-4 x+3} d x - \int \sqrt{x^2-4 x+3} d x \)
Let's solve the first integral, \( I_1 = \int(2 x-4) \sqrt{x^2-4 x+3} d x \).
Let \( u = x^2-4x+3 \). Then \( du = (2x-4) dx \).
So, \( I_1 = \int \sqrt{u} du = \int u^{1/2} du = \frac{u^{3/2}}{3/2} + C_1 = \frac{2}{3}(x^2-4x+3)^{3/2} + C_1 \).
Now, let's solve the second integral, \( I_2 = \int \sqrt{x^2-4 x+3} d x \).
Complete the square for \( x^2-4x+3 \):
\( x^2-4x+3 = x^2-4x+4-1 = (x-2)^2-1^2 \).
This is in the form \( \int \sqrt{t^2-a^2} dt \), where \( t=x-2 \) and \( a=1 \).
Using the formula \( \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( I_2 = \frac{(x-2) \sqrt{(x-2)^2-1}}{2} - \frac{1^2}{2}\log|x-2+\sqrt{(x-2)^2-1}| + C_2 \)
\( \implies = \frac{(x-2) \sqrt{x^2-4x+3}}{2} - \frac{1}{2}\log|x-2+\sqrt{x^2-4x+3}| + C_2 \).
Combine \( I_1 - I_2 \):
\( \int(2 x-5) \sqrt{x^2-4 x+3} d x = \frac{2}{3}(x^2-4x+3)^{3/2} - \left[ \frac{(x-2) \sqrt{x^2-4x+3}}{2} - \frac{1}{2}\log|x-2+\sqrt{x^2-4x+3}| \right] + C \)
\( \implies = \frac{2}{3}(x^2-4x+3)^{3/2} - \frac{(x-2) \sqrt{x^2-4x+3}}{2} + \frac{1}{2}\log|x-2+\sqrt{x^2-4x+3}| + C \)
In simple words: First, adjust \( (2x-5) \) so one part is the derivative of \( x^2-4x+3 \). Solve that part using simple substitution. For the other part, complete the square and use the correct integral formula involving the log function.

🎯 Exam Tip: This method works when the linear term \( Px+Q \) can be expressed as a multiple of the derivative of the quadratic part plus a constant. It neatly breaks down a complex integral into manageable standard forms.

Question 18. \( \int x \sqrt{x+x^2} d x \)
Answer: We will solve this integral by splitting it. The derivative of \( x^2+x \) is \( 2x+1 \). We need to manipulate the \( x \) outside the square root to get this term.
\( \int x \sqrt{x^2+x} d x = \int \frac{1}{2}(2x) \sqrt{x^2+x} d x \)
We need \( 2x+1 \), so we add and subtract 1 inside the bracket:
\( = \int \frac{1}{2}(2x+1-1) \sqrt{x^2+x} d x \)
\( \implies = \frac{1}{2} \int (2x+1) \sqrt{x^2+x} d x - \frac{1}{2} \int \sqrt{x^2+x} d x \)
Let's solve the first integral, \( I_1 = \frac{1}{2} \int (2x+1) \sqrt{x^2+x} d x \).
Let \( u = x^2+x \). Then \( du = (2x+1) dx \).
So, \( I_1 = \frac{1}{2} \int \sqrt{u} du = \frac{1}{2} \int u^{1/2} du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C_1 = \frac{1}{3}(x^2+x)^{3/2} + C_1 \).
Now, let's solve the second integral, \( I_2 = \frac{1}{2} \int \sqrt{x^2+x} d x \).
Complete the square for \( x^2+x \):
\( x^2+x = x^2+x+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \left(x+\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \).
This is in the form \( \int \sqrt{t^2-a^2} dt \), where \( t=x+\frac{1}{2} \) and \( a=\frac{1}{2} \).
Using the formula \( \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( I_2 = \frac{1}{2} \left[ \frac{\left(x+\frac{1}{2}\right) \sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}}{2} - \frac{\left(\frac{1}{2}\right)^2}{2}\log\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| \right] + C_2 \)
\( I_2 = \frac{1}{2} \left[ \frac{\frac{2x+1}{2} \sqrt{x^2+x}}{2} - \frac{\frac{1}{4}}{2}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C_2 \)
\( I_2 = \frac{1}{2} \left[ \frac{(2x+1) \sqrt{x^2+x}}{4} - \frac{1}{8}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C_2 \)
\( I_2 = \frac{(2x+1) \sqrt{x^2+x}}{8} - \frac{1}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| + C_2 \).
Combine \( I_1 - I_2 \):
\( \int x \sqrt{x^2+x} d x = \frac{1}{3}(x^2+x)^{3/2} - \left[ \frac{(2x+1) \sqrt{x^2+x}}{8} - \frac{1}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C \)
\( \implies = \frac{1}{3}(x^2+x)^{3/2} - \frac{(2x+1) \sqrt{x^2+x}}{8} + \frac{1}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| + C \)
In simple words: First, change \( x \) to make part of it the derivative of \( x^2+x \). Split the integral into two parts. Solve the first part with simple substitution. For the second part, complete the square and use the standard formula.

🎯 Exam Tip: For integrals like \( \int x \sqrt{ax^2+bx+c} dx \), rewrite \( x \) as \( A(2ax+b)+B \) to create a substitution-friendly term and a constant term that leads to completing the square.

Question 19. \( \int(x-5) \sqrt{x^2+x} d x \)
Answer: We will solve this integral by splitting it. The derivative of \( x^2+x \) is \( 2x+1 \). We need to manipulate \( (x-5) \) to include \( (2x+1) \).
We can write \( x-5 \) as \( \frac{1}{2}(2x+1-11) \).
So, \( \int(x-5) \sqrt{x^2+x} d x = \int \frac{1}{2}(2x+1-11) \sqrt{x^2+x} d x \)
\( \implies = \frac{1}{2} \int (2x+1) \sqrt{x^2+x} d x - \frac{11}{2} \int \sqrt{x^2+x} d x \)
Let's solve the first integral, \( I_1 = \frac{1}{2} \int (2x+1) \sqrt{x^2+x} d x \).
Let \( u = x^2+x \). Then \( du = (2x+1) dx \).
So, \( I_1 = \frac{1}{2} \int \sqrt{u} du = \frac{1}{2} \int u^{1/2} du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C_1 = \frac{1}{3}(x^2+x)^{3/2} + C_1 \).
Now, let's solve the second integral, \( I_2 = \frac{11}{2} \int \sqrt{x^2+x} d x \).
Complete the square for \( x^2+x \):
\( x^2+x = x^2+x+\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \left(x+\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \).
This is in the form \( \int \sqrt{t^2-a^2} dt \), where \( t=x+\frac{1}{2} \) and \( a=\frac{1}{2} \).
Using the formula \( \frac{x \sqrt{x^2-a^2}}{2} - \frac{a^2}{2}\log|x + \sqrt{x^2-a^2}| + C \):
\( I_2 = \frac{11}{2} \left[ \frac{\left(x+\frac{1}{2}\right) \sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}}{2} - \frac{\left(\frac{1}{2}\right)^2}{2}\log\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| \right] + C_2 \)
\( I_2 = \frac{11}{2} \left[ \frac{\frac{2x+1}{2} \sqrt{x^2+x}}{2} - \frac{\frac{1}{4}}{2}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C_2 \)
\( I_2 = \frac{11}{2} \left[ \frac{(2x+1) \sqrt{x^2+x}}{4} - \frac{1}{8}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C_2 \)
\( I_2 = \frac{11(2x+1) \sqrt{x^2+x}}{8} - \frac{11}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| + C_2 \).
Combine \( I_1 - I_2 \):
\( \int(x-5) \sqrt{x^2+x} d x = \frac{1}{3}(x^2+x)^{3/2} - \left[ \frac{11(2x+1) \sqrt{x^2+x}}{8} - \frac{11}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| \right] + C \)
\( \implies = \frac{1}{3}(x^2+x)^{3/2} - \frac{11(2x+1) \sqrt{x^2+x}}{8} + \frac{11}{16}\log\left|x+\frac{1}{2}+\sqrt{x^2+x}\right| + C \)
In simple words: This problem is solved by splitting the integral. Make \( x-5 \) into two parts: one that is the derivative of \( x^2+x \), and another constant part. Solve the first part by simple substitution, and the second part by completing the square and using the standard integral formula.

🎯 Exam Tip: When dealing with \( \int (Px+Q)\sqrt{quadratic} dx \), writing \( Px+Q = A \cdot (\text{derivative of quadratic}) + B \) is a reliable method. Calculate A and B by comparing coefficients.

Question 20. \( \int(x+3) \sqrt{3-4 x-x^2} d x \)
Answer: We solve this integral by splitting it. The derivative of \( 3-4x-x^2 \) is \( -4-2x \). We manipulate \( x+3 \) to include \( -4-2x \).
We can write \( x+3 \) as \( -\frac{1}{2}(-2x-6) = -\frac{1}{2}(-2x-4-2) \).
So, \( \int(x+3) \sqrt{3-4 x-x^2} d x = \int -\frac{1}{2}(-2x-4-2) \sqrt{3-4 x-x^2} d x \)
\( \implies = -\frac{1}{2} \int (-2x-4) \sqrt{3-4 x-x^2} d x - \frac{1}{2} \int (-2) \sqrt{3-4 x-x^2} d x \)
\( \implies = -\frac{1}{2} \int (-2x-4) \sqrt{3-4 x-x^2} d x + \int \sqrt{3-4 x-x^2} d x \)
Let's solve the first integral, \( I_1 = -\frac{1}{2} \int (-2x-4) \sqrt{3-4 x-x^2} d x \).
Let \( u = 3-4x-x^2 \). Then \( du = (-4-2x) dx \).
So, \( I_1 = -\frac{1}{2} \int \sqrt{u} du = -\frac{1}{2} \int u^{1/2} du = -\frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C_1 = -\frac{1}{3}(3-4x-x^2)^{3/2} + C_1 \).
Now, let's solve the second integral, \( I_2 = \int \sqrt{3-4 x-x^2} d x \).
Complete the square for \( 3-4x-x^2 \):
\( 3-4x-x^2 = 3 - (x^2+4x) = 3 - (x^2+4x+4-4) = 3 - ((x+2)^2-4) \)
\( \implies = 3 - (x+2)^2 + 4 = 7 - (x+2)^2 \).
We can write 7 as \( (\sqrt{7})^2 \).
So, the expression becomes \( (\sqrt{7})^2-(x+2)^2 \).
This is in the form \( \int \sqrt{a^2-t^2} dt \), where \( t=x+2 \) and \( a=\sqrt{7} \).
Using the formula \( \frac{x \sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \):
\( I_2 = \frac{(x+2) \sqrt{(\sqrt{7})^2-(x+2)^2}}{2} + \frac{(\sqrt{7})^2}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right) + C_2 \)
\( \implies = \frac{(x+2) \sqrt{7-(x+2)^2}}{2} + \frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right) + C_2 \)
\( \implies = \frac{(x+2) \sqrt{3-4x-x^2}}{2} + \frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right) + C_2 \).
Combine \( I_1 + I_2 \):
\( \int(x+3) \sqrt{3-4 x-x^2} d x = -\frac{1}{3}(3-4x-x^2)^{3/2} + \frac{(x+2) \sqrt{3-4x-x^2}}{2} + \frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right) + C \)
In simple words: Break this integral into two parts. Adjust \( x+3 \) so one part is the derivative of \( 3-4x-x^2 \) (remembering the minus signs), which can be solved with substitution. For the other part, complete the square to get the \( a^2-t^2 \) form, then use the standard integral formula involving \( \sin^{-1} \).

🎯 Exam Tip: Double-check the signs when completing the square, especially when starting with \( -x^2 \), as it impacts whether you get an \( a^2-x^2 \) or \( x^2-a^2 \) form, which determines the final formula.