Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.6

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6

Question 1. Evaluate \( \int \frac{x}{\sqrt{1+x^{2}}} \, dx \)
Answer:
To solve this integral, we use the substitution method.
Let \( 1 + x^2 = u \)
Now, differentiate both sides with respect to \( x \):
\( 2x \, dx = du \)

This means \( x \, dx = \frac{1}{2} du \)
Substitute these into the integral:
\( \int \frac{x}{\sqrt{1+x^2}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du \)
\( = \frac{1}{2} \int u^{-\frac{1}{2}} du \)
Now, we integrate using the power rule \( \int y^n dy = \frac{y^{n+1}}{n+1} \):
\( = \frac{1}{2} \left( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right) + c \)
\( = \frac{1}{2} \left( \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right) + c \)
\( = u^{\frac{1}{2}} + c \)
\( = \sqrt{u} + c \)
Finally, substitute back \( u = 1 + x^2 \):
\( = \sqrt{1+x^2} + c \)
In simple words: We change the variable inside the square root to 'u', then find what 'dx' becomes in terms of 'du'. This makes the integral simpler to solve using a basic power rule, and then we put the original 'x' back.

🎯 Exam Tip: For integrals involving expressions like \( \sqrt{f(x)} \) and \( f'(x) \cdot x \), try substituting \( f(x) = u \) or \( \sqrt{f(x)} = u \) to simplify the problem.

Question 2. Evaluate \( \int \frac{x^2}{1+x^6} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
First, rewrite the denominator: \( 1+x^6 = 1+(x^3)^2 \).
Let \( x^3 = u \)
Now, differentiate both sides with respect to \( x \):
\( 3x^2 \, dx = du \)

This gives us \( x^2 \, dx = \frac{1}{3} du \)
Substitute these into the integral:
\( \int \frac{x^2}{1+x^6} \, dx = \int \frac{1}{1+(x^3)^2} \cdot x^2 \, dx \)
\( = \int \frac{1}{1+u^2} \cdot \frac{1}{3} du \)
\( = \frac{1}{3} \int \frac{1}{1+u^2} du \)
We know that \( \int \frac{1}{1+y^2} dy = \tan^{-1}(y) + c \).
\( = \frac{1}{3} \tan^{-1}(u) + c \)
Finally, substitute back \( u = x^3 \):
\( = \frac{1}{3} \tan^{-1}(x^3) + c \)
In simple words: We see that the denominator has \( x^6 \), which is \( (x^3)^2 \). By letting \( x^3 \) be 'u', the integral becomes a standard form that leads to the inverse tangent function.

🎯 Exam Tip: Recognize patterns like \( 1+(f(x))^2 \) in the denominator and try substituting \( f(x) = u \) to use the inverse tangent integral formula.

Question 3. Evaluate \( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx \)
Answer:
To evaluate this integral, we will use the substitution method.
Let the denominator be \( u \):
Put \( e^x + e^{-x} = u \)
Now, differentiate both sides with respect to \( x \):
\( (e^x - e^{-x}) \, dx = du \)
Notice that the numerator is exactly this differential \( du \).
Substitute these into the integral:
\( \int \frac{e^x - e^{-x}}{e^x + e^{-x}} \, dx = \int \frac{du}{u} \)
We know that \( \int \frac{1}{y} dy = \log |y| + c \).
\( = \log |u| + c \)
Finally, substitute back \( u = e^x + e^{-x} \):
\( = \log |e^x + e^{-x}| + c \)
The absolute value is technically not needed here since \( e^x + e^{-x} \) is always positive.
In simple words: We pick the whole bottom part as 'u'. When we take its derivative, it perfectly matches the top part. So, the integral turns into a simple log function.

🎯 Exam Tip: If the numerator is the derivative of the denominator, the integral is simply the logarithm of the absolute value of the denominator.

Question 4. Evaluate \( \int \frac{10x^9 + 10^x \log_e 10}{10^x + x^{10}} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the denominator be \( u \):
Put \( 10^x + x^{10} = u \)
Now, differentiate both sides with respect to \( x \). Remember that the derivative of \( a^x \) is \( a^x \log_e a \) and the derivative of \( x^n \) is \( nx^{n-1} \):
\( (10^x \log_e 10 + 10x^9) \, dx = du \)
Notice that the numerator is exactly this differential \( du \).
Substitute these into the integral:
\( \int \frac{10x^9 + 10^x \log_e 10}{10^x + x^{10}} \, dx = \int \frac{du}{u} \)
\( = \log |u| + c \)
Finally, substitute back \( u = 10^x + x^{10} \):
\( = \log |10^x + x^{10}| + c \)
In simple words: This integral is special because the top part is the exact derivative of the bottom part. When this happens, the answer is always the natural logarithm of the bottom part.

🎯 Exam Tip: Always check if the numerator is the derivative of the denominator; this often simplifies the integral to a logarithm function.

Question 5. Evaluate \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let \( \sqrt{x} = u \)
Now, differentiate both sides with respect to \( x \). The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \):
\( \frac{1}{2\sqrt{x}} \, dx = du \)

We can rearrange this to get \( \frac{1}{\sqrt{x}} \, dx = 2 du \)
Substitute these into the integral:
\( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx = \int \sin u \cdot (2 du) \)
\( = 2 \int \sin u \, du \)
We know that \( \int \sin y \, dy = -\cos y + c \).
\( = 2 (-\cos u) + c \)
\( = -2 \cos u + c \)
Finally, substitute back \( u = \sqrt{x} \):
\( = -2 \cos \sqrt{x} + c \)
In simple words: We change the square root of x into 'u'. Then, we adjust 'dx' to match 'du'. This makes the problem a simple integral of sine, which we can solve easily.

🎯 Exam Tip: When you see \( \sqrt{x} \) in a trigonometric function, try substituting \( u = \sqrt{x} \) because its derivative \( \frac{1}{2\sqrt{x}} \) often appears elsewhere in the integrand.

Question 6. Evaluate \( \int \frac{\cot x}{\log (\sin x)} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the denominator (or part of it) be \( u \):
Put \( \log (\sin x) = u \)
Now, differentiate both sides with respect to \( x \). Remember to use the chain rule for \( \log (\sin x) \):
\( \frac{1}{\sin x} \cdot (\cos x) \, dx = du \)

We know that \( \frac{\cos x}{\sin x} = \cot x \), so:
\( \cot x \, dx = du \)
Substitute these into the integral:
\( \int \frac{\cot x}{\log (\sin x)} \, dx = \int \frac{1}{u} \, du \)
\( = \log |u| + c \)
Finally, substitute back \( u = \log (\sin x) \):
\( = \log |\log (\sin x)| + c \)
In simple words: We set the logarithm part as 'u'. When we find its derivative, it turns out to be exactly `cot x dx`, which is in the numerator. So, the integral becomes a straightforward logarithm.

🎯 Exam Tip: Look for nested functions like \( \log(f(x)) \). Often, substituting \( u = f(x) \) or \( u = \log(f(x)) \) will simplify the integral by making \( \frac{f'(x)}{f(x)} \) or \( \frac{1}{f(x)} f'(x) \) appear as part of `du`.

Question 7. Evaluate \( \int \frac{\operatorname{cosec} x}{\log \left(\tan \frac{x}{2}\right)} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the denominator be \( u \):
Put \( \log \left(\tan \frac{x}{2}\right) = u \)
Now, differentiate both sides with respect to \( x \). This requires the chain rule multiple times:
Derivative of \( \log(f(x)) \) is \( \frac{1}{f(x)} f'(x) \). Here \( f(x) = \tan \frac{x}{2} \).
Derivative of \( \tan(g(x)) \) is \( \sec^2(g(x)) g'(x) \). Here \( g(x) = \frac{x}{2} \).
So, \( \left( \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} \right) \, dx = du \)
Now, simplify the expression:
\( \left( \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} \right) \, dx = du \)
\( \left( \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \right) \, dx = du \)
Using the trigonometric identity \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \):
\( \frac{1}{\sin x} \, dx = du \)
We know that \( \frac{1}{\sin x} = \operatorname{cosec} x \), so:
\( \operatorname{cosec} x \, dx = du \)
Substitute these into the integral:
\( \int \frac{\operatorname{cosec} x}{\log \left(\tan \frac{x}{2}\right)} \, dx = \int \frac{1}{u} \, du \)
\( = \log |u| + c \)
Finally, substitute back \( u = \log \left(\tan \frac{x}{2}\right) \):
\( = \log \left|\log \left(\tan \frac{x}{2}\right)\right| + c \)
In simple words: We let the entire `log(tan(x/2))` be 'u'. When we find its derivative, it simplifies to `cosec x dx`. This transformation makes the integral a simple logarithm of 'u'.

🎯 Exam Tip: Integrals involving \( \log(\tan \frac{x}{2}) \) often simplify nicely because the derivative of \( \log(\tan \frac{x}{2}) \) is \( \operatorname{cosec} x \). Remember this useful identity.

Question 8. Evaluate \( \int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the denominator be \( u \):
Put \( a^2 + b^2 \sin^2 x = u \)
Now, differentiate both sides with respect to \( x \). Remember \( \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin 2x \):
\( (0 + b^2 \cdot 2 \sin x \cos x) \, dx = du \)

\( b^2 \sin 2x \, dx = du \)

From this, we can find \( \sin 2x \, dx \):
\( \sin 2x \, dx = \frac{1}{b^2} du \)
Substitute these into the integral:
\( \int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx = \int \frac{1}{u} \cdot \frac{1}{b^2} du \)
\( = \frac{1}{b^2} \int \frac{1}{u} du \)
\( = \frac{1}{b^2} \log |u| + c \)
Finally, substitute back \( u = a^2 + b^2 \sin^2 x \):
\( = \frac{1}{b^2} \log |a^2 + b^2 \sin^2 x| + c \)
In simple words: We let the whole denominator be 'u'. When we take its derivative, it involves `sin 2x`, which is in the numerator. This way, the integral simplifies to a basic log function with a constant factor.

🎯 Exam Tip: Always look for a function and its derivative within the integral. The identity \( \sin 2x = 2 \sin x \cos x \) is crucial when dealing with \( \sin^2 x \).

Question 9. Evaluate \( \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let \( \sin^{-1} x = u \)
Now, differentiate both sides with respect to \( x \). The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \):
\( \frac{1}{\sqrt{1-x^2}} \, dx = du \)
Substitute these into the integral:
\( \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx = \int u \, du \)
Now, we integrate using the power rule:
\( = \frac{u^{1+1}}{1+1} + c \)
\( = \frac{u^2}{2} + c \)
Finally, substitute back \( u = \sin^{-1} x \):
\( = \frac{(\sin^{-1} x)^2}{2} + c \)
In simple words: We choose the inverse sine function as 'u'. Its derivative is exactly the remaining part of the integral, making it a simple power rule integration problem.

🎯 Exam Tip: Whenever you see an inverse trigonometric function with its derivative present in the integral, substitution with the inverse function as 'u' is usually the key.

Question 10. Evaluate \( \int \frac{\sqrt{x}}{1+\sqrt{x}} \, dx \)
Answer:
To solve this integral, we will use the substitution method involving a change of variable.
Let \( 1+\sqrt{x} = u \)
From this, we can express \( \sqrt{x} \) and \( x \) in terms of \( u \):
\( \sqrt{x} = u - 1 \)
\( x = (u-1)^2 \)
Now, differentiate \( x = (u-1)^2 \) with respect to \( u \) to find \( dx \):
\( dx = 2(u-1) \cdot 1 \, du \)
Substitute these into the integral:
\( \int \frac{\sqrt{x}}{1+\sqrt{x}} \, dx = \int \frac{u-1}{u} \cdot 2(u-1) \, du \)
\( = 2 \int \frac{(u-1)^2}{u} \, du \)
Expand \( (u-1)^2 = u^2 - 2u + 1 \):
\( = 2 \int \frac{u^2 - 2u + 1}{u} \, du \)
Divide each term in the numerator by \( u \):
\( = 2 \int \left( \frac{u^2}{u} - \frac{2u}{u} + \frac{1}{u} \right) \, du \)
\( = 2 \int \left( u - 2 + \frac{1}{u} \right) \, du \)
Now, integrate each term:
\( = 2 \left[ \int u \, du - \int 2 \, du + \int \frac{1}{u} \, du \right] \)
\( = 2 \left[ \frac{u^2}{2} - 2u + \log |u| \right] + c \)
Finally, substitute back \( u = 1+\sqrt{x} \):
\( = 2 \left[ \frac{(1+\sqrt{x})^2}{2} - 2(1+\sqrt{x}) + \log |1+\sqrt{x}| \right] + c \)
\( = (1+\sqrt{x})^2 - 4(1+\sqrt{x}) + 2\log |1+\sqrt{x}| + c \)
In simple words: We make the whole denominator `(1+√x)` into 'u'. This also helps us change `√x` and `dx` into terms of 'u' and 'du'. After expanding and dividing, the integral becomes a mix of simple power rules and a logarithm, which we then integrate and substitute back.

🎯 Exam Tip: When an integral has a complex expression with \( \sqrt{x} \), substituting the entire expression including the square root, like \( u = 1+\sqrt{x} \), can often simplify the integrand significantly before you perform the integration.

Question 11. Evaluate \( \int \frac{1}{x \log x \log (\log x)} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the innermost logarithmic term be \( u \):
Put \( \log (\log x) = u \)
Now, differentiate both sides with respect to \( x \). Use the chain rule twice:
\( \frac{1}{\log x} \cdot \frac{1}{x} \, dx = du \)
Observe that \( \frac{1}{x \log x} \, dx \) is present in the original integral.
Substitute these into the integral:
\( \int \frac{1}{x \log x \log (\log x)} \, dx = \int \frac{1}{\log (\log x)} \cdot \left( \frac{1}{x \log x} \, dx \right) \)
\( = \int \frac{1}{u} \, du \)
\( = \log |u| + c \)
Finally, substitute back \( u = \log (\log x) \):
\( = \log |\log (\log x)| + c \)
In simple words: We choose the most deeply nested logarithm as 'u'. When we take its derivative, we find that the other parts of the integral conveniently form 'du'. This makes the integral very simple, resulting in a logarithm of 'u'.

🎯 Exam Tip: For nested logarithmic functions, substitute the innermost logarithm with 'u'. Its derivative will often "peel off" the outer layers, simplifying the integral to a basic form.

Question 12. Evaluate \( \int \alpha \beta x^{\alpha-1} e^{-\beta x^{\alpha}} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the exponent of \( e \) be \( u \):
Put \( -\beta x^{\alpha} = u \)
Now, differentiate both sides with respect to \( x \). Remember \( \frac{d}{dx}(x^\alpha) = \alpha x^{\alpha-1} \):
\( -\beta \alpha x^{\alpha-1} \, dx = du \)
We notice that the term \( \alpha \beta x^{\alpha-1} \, dx \) is present in the original integral. We can rearrange our derivative:
\( \alpha \beta x^{\alpha-1} \, dx = -du \)
Substitute these into the integral:
\( \int \alpha \beta x^{\alpha-1} e^{-\beta x^{\alpha}} \, dx = \int e^{u} (-du) \)
\( = - \int e^{u} du \)
We know that \( \int e^y dy = e^y + c \).
\( = -e^{u} + c \)
Finally, substitute back \( u = -\beta x^{\alpha} \):
\( = -e^{-\beta x^{\alpha}} + c \)
In simple words: We select the power of 'e' as 'u'. When we find its derivative, it perfectly matches the other `x` terms and constants outside the 'e' part, except for a sign. This simplifies the integral to a basic `e^u` form.

🎯 Exam Tip: When dealing with \( e^{f(x)} \), often substituting \( u = f(x) \) (or \( u = -f(x) \) if there's a negative sign) is effective, as \( f'(x) \, dx \) might be present elsewhere.

Question 13. Evaluate \( \int \tan x \sqrt{\sec x} \, dx \)
Answer:
To solve this integral, we will use the substitution method combined with trigonometric identities.
First, rewrite the integrand using basic trigonometric functions:
\( \tan x \sqrt{\sec x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\sqrt{\cos x}} = \frac{\sin x}{(\cos x)^{1 + \frac{1}{2}}} = \frac{\sin x}{(\cos x)^{\frac{3}{2}}} \)
So, the integral becomes \( \int \frac{\sin x}{(\cos x)^{\frac{3}{2}}} \, dx \)
Let \( \cos x = u \)
Now, differentiate both sides with respect to \( x \):
\( -\sin x \, dx = du \)

This implies \( \sin x \, dx = -du \)
Substitute these into the integral:
\( \int \frac{\sin x}{(\cos x)^{\frac{3}{2}}} \, dx = \int \frac{-du}{u^{\frac{3}{2}}} \)
\( = - \int u^{-\frac{3}{2}} du \)
Now, integrate using the power rule \( \int y^n dy = \frac{y^{n+1}}{n+1} \):
\( = - \left( \frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} \right) + c \)
\( = - \left( \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \right) + c \)
\( = 2 u^{-\frac{1}{2}} + c \)
\( = \frac{2}{\sqrt{u}} + c \)
Finally, substitute back \( u = \cos x \):
\( = \frac{2}{\sqrt{\cos x}} + c \)
This can also be written as \( 2 \sec^{\frac{1}{2}} x + c \) or \( 2\sqrt{\sec x} + c \).
In simple words: We first rewrite tangent and secant in terms of sine and cosine. Then, we substitute `cos x` with 'u'. Its derivative, `sin x dx`, helps simplify the top part, allowing us to solve the integral using a basic power rule.

🎯 Exam Tip: When dealing with products of trigonometric functions, especially with roots, converting them to sines and cosines often reveals a suitable substitution, like `u = cos x` or `u = sin x`.

Question 14. Evaluate \( \int x (1-x)^{17} \, dx \)
Answer:
To solve this integral, we will use the substitution method.
Let the term inside the parenthesis be \( u \):
Put \( 1-x = u \)
From this, we can express \( x \) in terms of \( u \):
\( x = 1-u \)
Now, differentiate \( 1-x = u \) with respect to \( x \):
\( -dx = du \)

This implies \( dx = -du \)
Substitute these into the integral:
\( \int x (1-x)^{17} \, dx = \int (1-u) u^{17} (-du) \)
\( = - \int (u^{17} - u^{18}) du \)
Now, separate the terms and integrate:
\( = - \left[ \int u^{17} du - \int u^{18} du \right] \)
\( = - \left[ \frac{u^{17+1}}{17+1} - \frac{u^{18+1}}{18+1} \right] + c \)
\( = - \left[ \frac{u^{18}}{18} - \frac{u^{19}}{19} \right] + c \)
\( = \frac{u^{19}}{19} - \frac{u^{18}}{18} + c \)
Finally, substitute back \( u = 1-x \):
\( = \frac{(1-x)^{19}}{19} - \frac{(1-x)^{18}}{18} + c \)
In simple words: We let `(1-x)` be 'u', which simplifies the power part. We also replace 'x' with `(1-u)` and `dx` with `-du`. This transforms the integral into a simpler polynomial, which is easy to integrate term by term.

🎯 Exam Tip: When you have a product like \( x(ax+b)^n \), substituting \( u = ax+b \) is very effective. Remember to express \( x \) in terms of \( u \) before substituting.

Question 15. Evaluate \( \int \sin^5 x \cos^3 x \, dx \)
Answer:
To solve this integral, we will use the substitution method along with trigonometric identities.
First, split the `cos^3 x` term to prepare for substitution:
\( \int \sin^5 x \cos^3 x \, dx = \int \sin^5 x \cos^2 x \cos x \, dx \)
Now, use the identity \( \cos^2 x = 1 - \sin^2 x \):
\( = \int \sin^5 x (1 - \sin^2 x) \cos x \, dx \)
Let \( \sin x = u \)
Now, differentiate both sides with respect to \( x \):
\( \cos x \, dx = du \)
Substitute these into the integral:
\( = \int u^5 (1 - u^2) \, du \)
Distribute \( u^5 \):
\( = \int (u^5 - u^7) \, du \)
Now, integrate each term using the power rule:
\( = \frac{u^{5+1}}{5+1} - \frac{u^{7+1}}{7+1} + c \)
\( = \frac{u^6}{6} - \frac{u^8}{8} + c \)
Finally, substitute back \( u = \sin x \):
\( = \frac{\sin^6 x}{6} - \frac{\sin^8 x}{8} + c \)
In simple words: We separate one `cos x` for `du` and convert the rest of the `cos^2 x` into `1 - sin^2 x`. Then, we replace `sin x` with 'u'. This turns the integral into a simple polynomial in 'u', which is easy to integrate.

🎯 Exam Tip: For integrals of powers of sine and cosine, if one power is odd, save one factor for `du` and convert the remaining even power using \( \sin^2 x + \cos^2 x = 1 \).

Question 16. Evaluate \( \int \frac{\cos x}{\cos (x-a)} \, dx \)
Answer:
To solve this integral, we will use the substitution method and trigonometric identities.
Let the argument of the denominator's cosine be \( u \):
Put \( x-a = u \)
From this, we can express \( x \) in terms of \( u \):
\( x = u+a \)
Now, differentiate \( x-a = u \) with respect to \( x \):
\( dx = du \)
Substitute these into the integral:
\( \int \frac{\cos x}{\cos (x-a)} \, dx = \int \frac{\cos (u+a)}{\cos u} \, du \)
Now, use the compound angle formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \):
\( = \int \frac{\cos u \cos a - \sin u \sin a}{\cos u} \, du \)
Separate the terms in the numerator:
\( = \int \left( \frac{\cos u \cos a}{\cos u} - \frac{\sin u \sin a}{\cos u} \right) \, du \)
\( = \int (\cos a - \tan u \sin a) \, du \)
Since \( a \) is a constant, \( \cos a \) and \( \sin a \) are also constants. Separate the integral into two parts:
\( = \cos a \int du - \sin a \int \tan u \, du \)
We know that \( \int du = u + C_1 \) and \( \int \tan u \, du = \log |\sec u| + C_2 \).
\( = \cos a (u) - \sin a \log |\sec u| + c \)
Finally, substitute back \( u = x-a \):
\( = \cos a (x-a) - \sin a \log |\sec (x-a)| + c \)
In simple words: We let `(x-a)` be 'u' so the denominator is simple. We then replace `x` in the numerator with `(u+a)` and expand it using the cosine addition formula. This allows us to break the integral into simpler parts, one with `cos a` and one with `tan u`, which we can then integrate.

🎯 Exam Tip: For integrals with trigonometric functions where the arguments are shifted (like `x-a`), substitute the shifted argument with 'u'. Then, use compound angle formulas to simplify the expression before integration.