Evaluate the integrals in Exercises 31-50 . Some integrals do not require integration by parts.

Evaluate the integrals in Exercises 31-50 . Some integrals...

Key Concepts

Reduction Formulas

Reduction formulas are systematic methods for expressing complex integrals in terms of simpler ones, typically by reducing the power of a polynomial factor in the integrand. They are especially useful when dealing with integrals that involve powers multiplied by trigonometric functions, enabling a step-by-step simplification.

Trigonometric Integrals

Trigonometric integrals focus on the integration of functions containing trigonometric terms, such as sine and cosine. Solving these integrals often involves techniques like substitution or integration by parts, particularly when the trigonometric function has a compound argument or is paired with an algebraic function.

Integration by Parts

Integration by parts is a fundamental technique for integrating the product of two types of functions. It uses the formula ?u dv = uv - ?v du, which is particularly effective when one function (often a polynomial) becomes simpler upon differentiation, while the other (often a trigonometric function) has an easily computable antiderivative.

Definite Integration

Definite integration involves evaluating the antiderivative of a function at specified upper and lower limits, resulting in the net accumulation or area under the curve over an interval. This process requires careful consideration of the boundary values during evaluation.

Transcript

00:01 Problem 48 says that we have to find the value of a definite integral 0 to pi over 2 x cubed, cos 2x x.

00:13 So this will be done by integration by parts.

00:15 So this is algebraic, so first term, trigon matrix, second term.

00:19 So this will become first term as it is integral of cost 2x will be sine 2x over 2 minus integration of differentiation of x cube will be 3x square.

00:31 Of cost 2x will be once again sine 2x over 2 and this should be integrated once again so this becomes x cube over 2 sine 2 x minus 3 over 2 integration of x squared sine 2 x dx once again integration by parts first term second term so this becomes x cube over 2 sine 2x minus 3 over 2 first term remains as it is.

01:02 Integral of sine 2x will be minus cos 2x over 2.

01:06 Minus integral of differentiation of x squared is 2x.

01:10 Integral of sine 2x again is minus cos 2x over 2.

01:16 So minus cos 2x over 2 and this should be integrated once again.

01:28 Let's rewrite this and open the brackets.

01:32 So we have sine 2x over here minus.

01:37 This will become plus 3 over 4 x square cost 2x then here we have three negatives so 1 2 and 3 so this will become minus of and this will become 3 times 2 6 over 4 and we have here we have the integral of cos 2x and for x cost 2 x x x cost 2 x x x cost 2 x and d x once again for the last time integration by parts first term and second term.

02:21 So we have a do over here minus 6 over 4.

02:24 First term is as it is integral of cost 2x will be sine 2x over 2.

02:32 Minus integral of differential of first term will be 1...

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