You will be able to
- Apply integration by parts to products such as x·sin x and ln x
- Simplify powers of sine and cosine with trig identities and u-substitution
- Integrate tangent and secant products by rewriting them with the identity 1 + tan²x = sec²x
- Use the half-angle identity to integrate sin²x
- Set up trigonometric substitutions for radicals of the form √(a² − x²) and evaluate the resulting integral
Good to know first
- Differentiation rules, including the product and chain rules
- Basic u-substitution for indefinite integrals
- Core trig identities such as sin²x + cos²x = 1 and the derivatives of sin, cos, and tan
- Right-triangle trig (SOH-CAH-TOA) and the Pythagorean theorem
Syllabus
7 lessons
- 1
Integration by parts: integrating x·sin x Free
You will learn the integration by parts formula, the integral of u dv = uv minus the integral of v du, and use it to integrate a product of a linear function and a trig function. You will see how choosing u and dv turns a hard integral into an easier one.
2 exercises3 questions - 2
Integration by parts: the integral of ln x Beyond Pass
You will learn how to choose u and dv for integration by parts, using the integral of ln x as the example. You will apply the formula ∫u dv = uv − ∫v du step by step and check the result.
5 build steps2 exercises4 questions - 3
Odd powers of cosine via u-substitution Beyond Pass
When cosine has an odd power, split off one cosine and turn the rest into sines using sin² + cos² = 1. Then the leftover cos x dx is exactly du for u = sin x, and the integral becomes a simple power-rule problem.
3 exercises3 questions - 4
Odd cosine and even sine powers Beyond Pass
When the cosine power is odd, you split off one cosine and turn the rest into sines, then use u-substitution with u = sin x. This lesson walks through the full integral of cos^5 x sin^4 x.
3 formulas2 exercises3 questions - 5
Integrating sin²x with the half-angle identity Beyond Pass
You will learn how to rewrite sin²x using the half-angle identity so it can be integrated directly, even though u-substitution does not work on it. You will also see why the integral of cos(2x) is (1/2)sin(2x), using u-substitution.
4 formulas3 exercises3 questions - 6
Tangent and secant integrals: splitting off a sec² and substituting Beyond Pass
You learn how to integrate a product of a tangent power and an even secant power by splitting off one sec²x to make du, then turning the leftover sec² into tan² with the identity 1 + tan²x = sec²x. The result is a polynomial in u that the power rule can integrate.
1 code5 build steps3 exercises3 questions - 7
Trigonometric substitution with √(4 − x²) Beyond Pass
You learn how to recognise a square root of the form √(a² − x²) and replace x with a sine of θ so the root becomes a simple cosine. You then turn the answer back into x using a right triangle.
2 exercises3 questions
Free preview · Lesson 1 in full
Integration by parts: integrating x·sin x
You will learn the integration by parts formula, the integral of u dv = uv minus the integral of v du, and use it to integrate a product of a linear function and a trig function. You will see how choosing u and dv turns a hard integral into an easier one.
Draw and write next to the video.
The idea
Some integrals are products of two different kinds of functions, such as x times sin x. A simple substitution does not work here, because the x and the sin x are multiplied together and cannot be separated. Integration by parts is a technique for exactly this situation. It rewrites one hard integral as a product minus a second integral that is usually easier.
The formula is . You split the integrand into two pieces: a piece called u, which you will differentiate, and a piece called dv, which you will integrate. Then you need du (the derivative of u times dx) and v (the integral of dv).
For , pick u = x, because its derivative is simple: du = dx. The rest, dv = sin x dx, is what you integrate. Integrating sin x gives v = −cos x. Notice you do not need a '+ C' yet; you add it once at the very end.
Now plug into the formula. The first term is u·v = x(−cos x) = −x cos x. The second term is −∫v du = −∫(−cos x) dx = +∫cos x dx. The integral of cos x is sin x, so the result is −x cos x + sin x + C.
Check your answer by differentiating. The derivative of −x cos x is −cos x + x sin x, and the derivative of sin x is cos x. Adding them gives x sin x, which is the original integrand. This check is a great habit to build.
Key points
- Integration by parts: . 0:36
- Choose u to be the part that gets simpler when you differentiate it; here u = x, so du = dx. 1:05
- dv is the rest of the integrand, here sin x dx, and v is its integral: v = −cos x. 1:25
- Then the integral becomes uv minus the integral of v du, which is −x cos x − ∫(−cos x) dx. 2:11
- Simplify the two negatives and integrate cos x to get the final answer −x cos x + sin x + C. 2:39
Practice
Use integration by parts to find . Choose u = x and dv = cos x dx. What is the value of the antiderivative at x = 0 when you take the constant C = 0? Give the value of x sin x + cos x at x = 0 (that is, the result of the antiderivative with C = 0 at x = 0).
Hint
Find v = sin x. Then the integral is x sin x − ∫sin x dx = x sin x + cos x + C. Evaluate at x = 0 with C = 0.
Solution
- Let u = x, so du = dx. Let dv = cos x dx, so v = sin x.
- Apply the formula: ∫x cos x dx = x sin x − ∫sin x dx.
- ∫sin x dx = −cos x, so the result is x sin x + cos x + C.
- At x = 0 with C = 0: 0·sin 0 + cos 0 = 0 + 1 = 1.
1
In your own words, explain why we chose u = x instead of u = sin x in the example, and what would happen to the integral if we had chosen differently.
Hint
Think about what happens to each piece when you differentiate or integrate it.
Solution
- Choosing u = x works because differentiating x gives 1, which makes the new integral simpler.
- If we had chosen u = sin x, its derivative cos x would still be a trig function, and integrating dv = x dx gives x²/2, so the new integral would be more complicated, not simpler.
Choosing u = x makes du = dx, so the leftover integral ∫v du becomes the simple ∫cos x dx. Choosing the part that gets simpler when differentiated is the key idea.
On your scratchpad
Draw a box for each piece of ∫ x sin x dx: label one box u and the other dv. Then write u, du, dv and v beside them and fill in the formula uv − ∫v du.
Quiz
In , which choice does the video make for u?
- u = sin x
- u = x
- u = x sin x
- u = dx
Show the answer
B. u = x · The video sets u = x, so du = dx, and dv = sin x dx.
1:05 watch the momentWhat is the integral of v du after choosing v = −cos x and du = dx?
- ∫ cos x dx
- ∫ −cos x dx
- ∫ x dx
- ∫ sin x dx
Show the answer
A. ∫ cos x dx · −∫ v du = −∫(−cos x)dx = +∫cos x dx, because the two negatives cancel.
2:11 watch the momentWhat is the final answer to ∫ x sin x dx?
- x cos x + sin x + C
- −x cos x + sin x + C
- −x cos x − sin x + C
- x sin x + C
Show the answer
B. −x cos x + sin x + C · The formula gives −x cos x + sin x + C.
2:41 watch the moment
Lessons 2–7
They open with Beyond Pass: every lesson of every course, in full. One payment: $14.99 for 90 days or $29 for a year. No subscription.
Made by Beyond from the video “Calculus 2 - Basic Integration” by The Organic Chemistry Tutor. The video plays in YouTube's own player. Watch it on YouTube · All lessons