OP Malhotra Class 12 Maths Solutions Chapter 14 Indefinite Integral 2 Exercise 14 (D)

Question 1. \( \int e^x \sin x d x \)

Answer: Let \( I = \int e^x \sin x \, dx \). We will solve this integral using integration by parts.
First, we apply integration by parts. Let \( u = \sin x \) and \( dv = e^x \, dx \).
Then \( du = \cos x \, dx \) and \( v = e^x \).
Using the formula \( \int u \, dv = uv - \int v \, du \), we get:
\( I = e^x \sin x - \int e^x \cos x \, dx \) --- (1)
Now, we need to integrate \( \int e^x \cos x \, dx \) using integration by parts again.
Let \( u = \cos x \) and \( dv = e^x \, dx \).
Then \( du = -\sin x \, dx \) and \( v = e^x \).
So, \( \int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx \)
Substitute this result back into equation (1):
\( I = e^x \sin x - (e^x \cos x + \int e^x \sin x \, dx) \)
\( \implies I = e^x \sin x - e^x \cos x - I \)
\( \implies 2I = e^x (\sin x - \cos x) \)
\( \implies I = \frac{e^x}{2} (\sin x - \cos x) + C \)
This type of integral is often called a recurrent integral because the original integral appears again in its own solution, allowing you to solve for it.
In simple words: To solve this integral, we use a method called integration by parts two times. After doing it twice, the original integral comes back in the equation. Then, we can simply solve for the integral as if it were a normal variable.

🎯 Exam Tip: Remember the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to choose 'u' for integration by parts. For integrals like \( \int e^{ax} \sin(bx) \, dx \) or \( \int e^{ax} \cos(bx) \, dx \), you will always need to apply integration by parts twice.

Question 2. \( \int e^{2 x} \sin 3 x d x \)

Answer: Let \( I = \int e^{2 x} \sin 3 x \, dx \). We will solve this using integration by parts, applying it twice.
First, we use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Let \( u = \sin 3x \) and \( dv = e^{2x} \, dx \).
Then, \( du = 3 \cos 3x \, dx \) and \( v = \frac{e^{2x}}{2} \).
So, we get:
\( I = \frac{e^{2x}}{2} \sin 3x - \int \frac{e^{2x}}{2} (3 \cos 3x) \, dx \)
\( \implies I = \frac{e^{2x} \sin 3x}{2} - \frac{3}{2} \int e^{2x} \cos 3x \, dx \)
Now, we apply integration by parts again to the new integral, \( \int e^{2x} \cos 3x \, dx \).
Let \( u = \cos 3x \) and \( dv = e^{2x} \, dx \).
Then, \( du = -3 \sin 3x \, dx \) and \( v = \frac{e^{2x}}{2} \).
This gives us:
\( \int e^{2x} \cos 3x \, dx = \frac{e^{2x}}{2} \cos 3x - \int \frac{e^{2x}}{2} (-3 \sin 3x) \, dx \)
\( = \frac{e^{2x}}{2} \cos 3x + \frac{3}{2} \int e^{2x} \sin 3x \, dx \)
Substitute this result back into the expression for \( I \):
\( I = \frac{e^{2x} \sin 3x}{2} - \frac{3}{2} \left[ \frac{e^{2x}}{2} \cos 3x + \frac{3}{2} \int e^{2x} \sin 3x \, dx \right] \)
\( \implies I = \frac{e^{2x} \sin 3x}{2} - \frac{3e^{2x} \cos 3x}{4} - \frac{9}{4} \int e^{2x} \sin 3x \, dx \)
Notice that the original integral \( I \) appeared again. We can now solve for \( I \):
\( \implies I + \frac{9}{4} I = \frac{e^{2x} \sin 3x}{2} - \frac{3e^{2x} \cos 3x}{4} \)
\( \implies \frac{13}{4} I = \frac{2e^{2x} \sin 3x - 3e^{2x} \cos 3x}{4} \)
\( \implies \frac{13}{4} I = \frac{e^{2x}}{4} (2 \sin 3x - 3 \cos 3x) \)
Multiply both sides by \( \frac{4}{13} \) to find \( I \):
\( \implies I = \frac{e^{2x}}{13} (2 \sin 3x - 3 \cos 3x) + C \)
In simple words: This integral is solved by using integration by parts two times. You first integrate part of the function, and then you do it again on the new integral that appears. Eventually, the original integral comes back, which lets you collect similar terms and find the answer.

🎯 Exam Tip: For integrals of the form \( \int e^{ax} \sin(bx) \, dx \) or \( \int e^{ax} \cos(bx) \, dx \), you can also remember the general formula to quickly check your answer, but showing all integration by parts steps is crucial for full marks.

Question 3. \( \int e^{-x} \sin x d x \)

Answer: Let \( I = \int e^{-x} \sin x \, dx \). We will solve this integral by applying integration by parts twice, similar to previous problems of this type.
First, we use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Let \( u = \sin x \) and \( dv = e^{-x} \, dx \).
Then, \( du = \cos x \, dx \) and \( v = -e^{-x} \).
So, we get:
\( I = (\sin x)(-e^{-x}) - \int (-e^{-x}) (\cos x) \, dx \)
\( \implies I = -e^{-x} \sin x + \int e^{-x} \cos x \, dx \) --- (1)
Now, we apply integration by parts again to the new integral, \( \int e^{-x} \cos x \, dx \).
Let \( u = \cos x \) and \( dv = e^{-x} \, dx \).
Then, \( du = -\sin x \, dx \) and \( v = -e^{-x} \).
This gives us:
\( \int e^{-x} \cos x \, dx = (\cos x)(-e^{-x}) - \int (-e^{-x}) (-\sin x) \, dx \)
\( = -e^{-x} \cos x - \int e^{-x} \sin x \, dx \)
Substitute this result back into equation (1):
\( I = -e^{-x} \sin x + \left[ -e^{-x} \cos x - \int e^{-x} \sin x \, dx \right] \)
\( \implies I = -e^{-x} \sin x - e^{-x} \cos x - I \)
\( \implies 2I = -e^{-x} (\sin x + \cos x) \)
\( \implies I = -\frac{e^{-x}}{2} (\sin x + \cos x) + C \)
In such recurrent integrals, the constant of integration from the second integration by parts is absorbed into the final constant C.
In simple words: This problem is solved by using the integration by parts method two times in a row. When the integral you started with shows up again, you can move it to the other side of the equation and solve for it easily.

🎯 Exam Tip: Be extra careful with the negative signs when integrating \( e^{-x} \) and when dealing with \( du \) for trigonometric functions. A common mistake is to forget the chain rule derivative or integral sign.

Question 4. (i) \( \int \cos (\log x) d x \)
(ii) \( \int \sin (\log x) d x \)

Answer:
(i) Let \( I = \int \cos(\log x) \, dx \). We use substitution to make this integral easier.
Let \( \log x = t \).
\( \implies x = e^t \)
Now, differentiate both sides with respect to \( t \): \( dx = e^t \, dt \).
Substitute these into the integral:
\( I = \int \cos t \cdot e^t \, dt \)
We solve this using integration by parts, which needs to be applied twice. Let \( u = \cos t \) and \( dv = e^t \, dt \).
Then \( du = -\sin t \, dt \) and \( v = e^t \).
\( I = e^t \cos t - \int e^t (-\sin t) \, dt \)
\( \implies I = e^t \cos t + \int e^t \sin t \, dt \) --- (A)
Now, we integrate \( \int e^t \sin t \, dt \) by parts again. Let \( u = \sin t \) and \( dv = e^t \, dt \).
Then \( du = \cos t \, dt \) and \( v = e^t \).
\( \int e^t \sin t \, dt = e^t \sin t - \int e^t \cos t \, dt \)
Substitute this back into equation (A):
\( I = e^t \cos t + (e^t \sin t - \int e^t \cos t \, dt) \)
\( \implies I = e^t \cos t + e^t \sin t - I \)
\( \implies 2I = e^t (\cos t + \sin t) \)
\( \implies I = \frac{e^t}{2} (\cos t + \sin t) + C \)
Now, substitute back \( t = \log x \):
\( I = \frac{e^{\log x}}{2} (\cos(\log x) + \sin(\log x)) + C \)
Since \( e^{\log x} = x \), the final answer is:
\( I = \frac{x}{2} (\cos(\log x) + \sin(\log x)) + C \)
(ii) Let \( I = \int \sin(\log x) \, dx \). Again, we use substitution to simplify the integral.
Let \( \log x = t \).
\( \implies x = e^t \)
Then \( dx = e^t \, dt \).
Substitute these into the integral:
\( I = \int \sin t \cdot e^t \, dt \)
We solve this using integration by parts, applying it twice. Let \( u = \sin t \) and \( dv = e^t \, dt \).
Then \( du = \cos t \, dt \) and \( v = e^t \).
\( I = e^t \sin t - \int e^t \cos t \, dt \) --- (B)
Now, we integrate \( \int e^t \cos t \, dt \) by parts again. Let \( u = \cos t \) and \( dv = e^t \, dt \).
Then \( du = -\sin t \, dt \) and \( v = e^t \).
\( \int e^t \cos t \, dt = e^t \cos t - \int e^t (-\sin t) \, dt \)
\( = e^t \cos t + \int e^t \sin t \, dt \)
Substitute this back into equation (B):
\( I = e^t \sin t - (e^t \cos t + \int e^t \sin t \, dt) \)
\( \implies I = e^t \sin t - e^t \cos t - I \)
\( \implies 2I = e^t (\sin t - \cos t) \)
\( \implies I = \frac{e^t}{2} (\sin t - \cos t) + C \)
Now, substitute back \( t = \log x \):
\( I = \frac{e^{\log x}}{2} (\sin(\log x) - \cos(\log x)) + C \)
Since \( e^{\log x} = x \), the final answer is:
\( I = \frac{x}{2} (\sin(\log x) - \cos(\log x)) + C \)
In simple words: For both parts, first change the variable from \( x \) to \( t \) by letting \( \log x = t \). This makes the integral look like the ones we just solved with \( e^t \) and sin/cos \( t \). Then, use integration by parts twice to solve for the integral, and finally change \( t \) back to \( \log x \).

🎯 Exam Tip: Remember the substitution \( \log x = t \implies x = e^t \implies dx = e^t \, dt \). This is a standard trick for integrals involving \( \log x \) and makes them solvable by methods used for \( \int e^{ax} \sin(bx) \, dx \) type integrals.

Question 5. \( \int e^{2 x} \sin x \cos x d x \)

Answer: Let \( I = \int e^{2 x} \sin x \cos x \, dx \). First, we simplify the trigonometric part.
We know that \( 2 \sin x \cos x = \sin 2x \). So, \( \sin x \cos x = \frac{1}{2} \sin 2x \).
Substitute this into the integral:
\( I = \int e^{2 x} \left( \frac{1}{2} \sin 2x \right) \, dx \)
\( \implies I = \frac{1}{2} \int e^{2 x} \sin 2x \, dx \)
Now, we solve the integral \( \int e^{2 x} \sin 2x \, dx \) using integration by parts twice. Let this integral be \( J \).
For \( J = \int e^{2 x} \sin 2x \, dx \), let \( u = \sin 2x \) and \( dv = e^{2x} \, dx \).
Then, \( du = 2 \cos 2x \, dx \) and \( v = \frac{e^{2x}}{2} \).
Applying integration by parts:
\( J = \frac{e^{2x}}{2} \sin 2x - \int \frac{e^{2x}}{2} (2 \cos 2x) \, dx \)
\( \implies J = \frac{e^{2x}}{2} \sin 2x - \int e^{2x} \cos 2x \, dx \) --- (A)
Next, integrate \( \int e^{2x} \cos 2x \, dx \) by parts. Let \( u = \cos 2x \) and \( dv = e^{2x} \, dx \).
Then, \( du = -2 \sin 2x \, dx \) and \( v = \frac{e^{2x}}{2} \).
\( \int e^{2x} \cos 2x \, dx = \frac{e^{2x}}{2} \cos 2x - \int \frac{e^{2x}}{2} (-2 \sin 2x) \, dx \)
\( = \frac{e^{2x}}{2} \cos 2x + \int e^{2x} \sin 2x \, dx \)
Notice that \( \int e^{2x} \sin 2x \, dx \) is our original integral \( J \). Substitute this back into equation (A):
\( J = \frac{e^{2x}}{2} \sin 2x - \left( \frac{e^{2x}}{2} \cos 2x + J \right) \)
\( \implies J = \frac{e^{2x}}{2} \sin 2x - \frac{e^{2x}}{2} \cos 2x - J \)
\( \implies 2J = \frac{e^{2x}}{2} (\sin 2x - \cos 2x) \)
\( \implies J = \frac{e^{2x}}{4} (\sin 2x - \cos 2x) + C_1 \)
Finally, substitute \( J \) back into the expression for \( I \):
\( I = \frac{1}{2} J \)
\( I = \frac{1}{2} \left[ \frac{e^{2x}}{4} (\sin 2x - \cos 2x) \right] + C \)
\( \implies I = \frac{e^{2x}}{8} (\sin 2x - \cos 2x) + C \)
This problem shows how trigonometric identities can simplify integrals before applying complex methods.
In simple words: First, change the `sin x cos x` part into `(1/2) sin 2x` using a special math rule. Then, you solve the new integral by doing integration by parts two times, just like in previous problems. Finally, remember to multiply the answer by the `1/2` you took out at the beginning.

🎯 Exam Tip: Always look for trigonometric identities like \( \sin x \cos x = \frac{1}{2} \sin 2x \), \( \sin^2 x = \frac{1 - \cos 2x}{2} \), or \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to simplify integrals before beginning complex integration methods.

Question 6. \( \int e^x \sin^2 x d x \)

Answer: Let \( I = \int e^x \sin^2 x \, dx \). We start by using a trigonometric identity to simplify the integral.
We know that \( \sin^2 x = \frac{1 - \cos 2x}{2} \).
Substitute this into the integral:
\( I = \int e^x \left( \frac{1 - \cos 2x}{2} \right) \, dx \)
\( \implies I = \frac{1}{2} \int e^x \, dx - \frac{1}{2} \int e^x \cos 2x \, dx \)
The first part is simple: \( \frac{1}{2} \int e^x \, dx = \frac{1}{2} e^x \).
Now, let's solve the second integral, \( J = \int e^x \cos 2x \, dx \), using integration by parts twice.
For \( J \), let \( u = \cos 2x \) and \( dv = e^x \, dx \).
Then \( du = -2 \sin 2x \, dx \) and \( v = e^x \).
Applying integration by parts:
\( J = e^x \cos 2x - \int e^x (-2 \sin 2x) \, dx \)
\( \implies J = e^x \cos 2x + 2 \int e^x \sin 2x \, dx \) --- (A)
Next, integrate \( \int e^x \sin 2x \, dx \) by parts. Let \( u = \sin 2x \) and \( dv = e^x \, dx \).
Then \( du = 2 \cos 2x \, dx \) and \( v = e^x \).
\( \int e^x \sin 2x \, dx = e^x \sin 2x - \int e^x (2 \cos 2x) \, dx \)
\( = e^x \sin 2x - 2 \int e^x \cos 2x \, dx \)
Notice that \( \int e^x \cos 2x \, dx \) is our original integral \( J \). Substitute this back into equation (A):
\( J = e^x \cos 2x + 2 (e^x \sin 2x - 2J) \)
\( \implies J = e^x \cos 2x + 2e^x \sin 2x - 4J \)
\( \implies 5J = e^x (\cos 2x + 2 \sin 2x) \)
\( \implies J = \frac{e^x}{5} (\cos 2x + 2 \sin 2x) \)
Now, substitute \( J \) back into the expression for \( I \):
\( I = \frac{1}{2} e^x - \frac{1}{2} \left[ \frac{e^x}{5} (\cos 2x + 2 \sin 2x) \right] + C \)
\( \implies I = \frac{e^x}{2} - \frac{e^x}{10} (\cos 2x + 2 \sin 2x) + C \)
\( \implies I = \frac{5e^x - e^x (\cos 2x + 2 \sin 2x)}{10} + C \)
\( \implies I = \frac{e^x}{10} (5 - \cos 2x - 2 \sin 2x) + C \)
This method involves simplifying a trigonometric function first before performing a recurrent integration.
In simple words: First, you change `sin^2 x` into a different form using a trig identity. This splits the integral into two parts. One part is easy to solve, and the other part needs integration by parts done twice. After solving both, combine them for the final answer.

🎯 Exam Tip: Always look for ways to simplify the integrand using trigonometric identities like \( \sin^2 x = \frac{1 - \cos 2x}{2} \) or \( \cos^2 x = \frac{1 + \cos 2x}{2} \) before attempting integration, as it can significantly reduce complexity.

Question 7. \( \int \frac{1}{x^3} \sin (\log x) d x \)

Answer: Let \( I = \int \frac{1}{x^3} \sin (\log x) \, dx \). We begin by using a substitution to simplify the integral.
Let \( \log x = t \).
\( \implies x = e^t \)
Then, \( dx = e^t \, dt \).
Also, \( \frac{1}{x^3} = x^{-3} = (e^t)^{-3} = e^{-3t} \).
Substitute these into the integral:
\( I = \int e^{-3t} \sin t \cdot e^t \, dt \)
\( \implies I = \int e^{-2t} \sin t \, dt \)
Now, we solve this integral using integration by parts, applying it twice. Let \( u = \sin t \) and \( dv = e^{-2t} \, dt \).
Then, \( du = \cos t \, dt \) and \( v = -\frac{e^{-2t}}{2} \).
Applying integration by parts:
\( I = (\sin t) \left(-\frac{e^{-2t}}{2}\right) - \int \left(-\frac{e^{-2t}}{2}\right) (\cos t) \, dt \)
\( \implies I = -\frac{e^{-2t}}{2} \sin t + \frac{1}{2} \int e^{-2t} \cos t \, dt \) --- (A)
Next, integrate \( \int e^{-2t} \cos t \, dt \) by parts. Let \( u = \cos t \) and \( dv = e^{-2t} \, dt \).
Then, \( du = -\sin t \, dt \) and \( v = -\frac{e^{-2t}}{2} \).
\( \int e^{-2t} \cos t \, dt = (\cos t) \left(-\frac{e^{-2t}}{2}\right) - \int \left(-\frac{e^{-2t}}{2}\right) (-\sin t) \, dt \)
\( = -\frac{e^{-2t}}{2} \cos t - \frac{1}{2} \int e^{-2t} \sin t \, dt \)
Notice that \( \int e^{-2t} \sin t \, dt \) is our original integral \( I \). Substitute this back into equation (A):
\( I = -\frac{e^{-2t}}{2} \sin t + \frac{1}{2} \left[ -\frac{e^{-2t}}{2} \cos t - \frac{1}{2} I \right] \)
\( \implies I = -\frac{e^{-2t}}{2} \sin t - \frac{e^{-2t}}{4} \cos t - \frac{1}{4} I \)
Now, bring the \( I \) terms together:
\( \implies I + \frac{1}{4} I = -\frac{e^{-2t}}{4} (2 \sin t + \cos t) \)
\( \implies \frac{5}{4} I = -\frac{e^{-2t}}{4} (2 \sin t + \cos t) \)
Multiply both sides by \( \frac{4}{5} \):
\( \implies I = -\frac{1}{5} e^{-2t} (2 \sin t + \cos t) + C \)
Finally, substitute back \( t = \log x \). Remember that \( e^{-2t} = e^{-2 \log x} = e^{\log x^{-2}} = x^{-2} = \frac{1}{x^2} \).
\( I = -\frac{1}{5x^2} (2 \sin(\log x) + \cos(\log x)) + C \)
This problem combines substitution with repeated integration by parts, a common technique for such integrals.
In simple words: First, replace `log x` with `t` and `dx` with `e^t dt`. This also changes `1/x^3` to `e^-3t`. The integral simplifies to `e^-2t sin t`. Then, solve this new integral using integration by parts two times, similar to earlier problems. Finally, change `t` back to `log x` to get the answer in terms of `x`.

🎯 Exam Tip: When using substitution for \( \log x = t \), remember to adjust the `x` terms in the integrand, such as \( x^{-3} \), to terms of \( t \) like \( e^{-3t} \), and also change \( dx \) to \( e^t \, dt \).

Question 8. \( \int e^{2 x} \cos^2 x d x \)

Answer: Let \( I = \int e^{2 x} \cos^2 x \, dx \). We start by using a trigonometric identity to simplify the integral.
We know that \( \cos^2 x = \frac{1 + \cos 2x}{2} \).
Substitute this into the integral:
\( I = \int e^{2 x} \left( \frac{1 + \cos 2x}{2} \right) \, dx \)
\( \implies I = \frac{1}{2} \int e^{2 x} \, dx + \frac{1}{2} \int e^{2 x} \cos 2x \, dx \)
The first integral is: \( \frac{1}{2} \int e^{2 x} \, dx = \frac{1}{2} \cdot \frac{e^{2x}}{2} = \frac{e^{2x}}{4} \).
Now, let's solve the second integral, \( J = \int e^{2 x} \cos 2x \, dx \), using integration by parts twice.
For \( J \), let \( u = \cos 2x \) and \( dv = e^{2x} \, dx \).
Then \( du = -2 \sin 2x \, dx \) and \( v = \frac{e^{2x}}{2} \).
Applying integration by parts:
\( J = \frac{e^{2x}}{2} \cos 2x - \int \frac{e^{2x}}{2} (-2 \sin 2x) \, dx \)
\( \implies J = \frac{e^{2x}}{2} \cos 2x + \int e^{2x} \sin 2x \, dx \) --- (A)
Next, integrate \( \int e^{2x} \sin 2x \, dx \) by parts. Let \( u = \sin 2x \) and \( dv = e^{2x} \, dx \).
Then \( du = 2 \cos 2x \, dx \) and \( v = \frac{e^{2x}}{2} \).
\( \int e^{2x} \sin 2x \, dx = \frac{e^{2x}}{2} \sin 2x - \int \frac{e^{2x}}{2} (2 \cos 2x) \, dx \)
\( = \frac{e^{2x}}{2} \sin 2x - \int e^{2x} \cos 2x \, dx \)
Notice that \( \int e^{2x} \cos 2x \, dx \) is our original integral \( J \). Substitute this back into equation (A):
\( J = \frac{e^{2x}}{2} \cos 2x + \left( \frac{e^{2x}}{2} \sin 2x - J \right) \)
\( \implies J = \frac{e^{2x}}{2} \cos 2x + \frac{e^{2x}}{2} \sin 2x - J \)
\( \implies 2J = \frac{e^{2x}}{2} (\cos 2x + \sin 2x) \)
\( \implies J = \frac{e^{2x}}{4} (\cos 2x + \sin 2x) \)
Now, substitute \( J \) back into the expression for \( I \):
\( I = \frac{e^{2x}}{4} + \frac{1}{2} \left[ \frac{e^{2x}}{4} (\cos 2x + \sin 2x) \right] + C \)
\( \implies I = \frac{e^{2x}}{4} + \frac{e^{2x}}{8} (\cos 2x + \sin 2x) + C \)
\( \implies I = \frac{2e^{2x} + e^{2x} (\cos 2x + \sin 2x)}{8} + C \)
\( \implies I = \frac{e^{2x}}{8} (2 + \cos 2x + \sin 2x) + C \)
This problem shows how a trigonometric identity helps break down a complex integral into simpler parts for easier solution.
In simple words: First, rewrite `cos^2 x` using a trigonometric rule to split the integral into two parts. One part is easy, and the other needs integration by parts twice. Solve each part and then combine them for the final answer.

🎯 Exam Tip: Always be on the lookout for trigonometric identities that can transform squared or product terms into linear terms, simplifying the path to integration. \( \cos^2 x = \frac{1 + \cos 2x}{2} \) is a very useful identity for this purpose.

Question 9. \( \int x^2 e^{x^3} \cos x^3 dx \)

Answer: Let \( I = \int x^2 e^{x^3} \cos x^3 \, dx \). We start by using a substitution to simplify the integral.
Let \( t = x^3 \).
Then, differentiate both sides with respect to \( x \): \( dt = 3x^2 \, dx \).
This means \( x^2 \, dx = \frac{1}{3} dt \).
Substitute these into the integral:
\( I = \int e^t \cos t \left( \frac{1}{3} dt \right) \)
\( \implies I = \frac{1}{3} \int e^t \cos t \, dt \)
Now, we solve the integral \( J = \int e^t \cos t \, dt \) using integration by parts twice.
For \( J \), let \( u = \cos t \) and \( dv = e^t \, dt \).
Then, \( du = -\sin t \, dt \) and \( v = e^t \).
Applying integration by parts:
\( J = e^t \cos t - \int e^t (-\sin t) \, dt \)
\( \implies J = e^t \cos t + \int e^t \sin t \, dt \) --- (A)
Next, integrate \( \int e^t \sin t \, dt \) by parts. Let \( u = \sin t \) and \( dv = e^t \, dt \).
Then \( du = \cos t \, dt \) and \( v = e^t \).
\( \int e^t \sin t \, dt = e^t \sin t - \int e^t \cos t \, dt \)
Notice that \( \int e^t \cos t \, dt \) is our integral \( J \). Substitute this back into equation (A):
\( J = e^t \cos t + (e^t \sin t - J) \)
\( \implies J = e^t \cos t + e^t \sin t - J \)
\( \implies 2J = e^t (\cos t + \sin t) \)
\( \implies J = \frac{e^t}{2} (\cos t + \sin t) + C_1 \)
Finally, substitute \( J \) back into the expression for \( I \):
\( I = \frac{1}{3} J \)
\( I = \frac{1}{3} \left[ \frac{e^t}{2} (\cos t + \sin t) \right] + C \)
\( \implies I = \frac{e^t}{6} (\cos t + \sin t) + C \)
Now, substitute back \( t = x^3 \):
\( I = \frac{e^{x^3}}{6} (\cos x^3 + \sin x^3) + C \)
This integral shows how a clever substitution can transform an intimidating integral into a more familiar and solvable form.
In simple words: First, replace `x^3` with `t`. This also means `x^2 dx` becomes `(1/3) dt`. The integral then looks like `(1/3) ∫e^t cos t dt`. Solve the `∫e^t cos t dt` part using integration by parts twice, then multiply the result by `1/3`, and finally change `t` back to `x^3`.

🎯 Exam Tip: For integrals containing composite functions like \( e^{x^3} \) and \( \cos x^3 \) where \( x^2 \) is also present, substitution of the inner function (e.g., \( t = x^3 \)) is usually the first and most effective step.