part C(i) and (iii) (c) Observe that the examples in (b) work nicely because of the derivatives you wereasked to calculate in (a). Each integrand in (b) is precisely the result of differentiatingone of the products of basic functions found in (a). To see what happens when anintegrand is still a product but not necessarily the result of differentiating anelementary product, we consider how to evaluatex cos(x) dx.(1) First, observe thatd[x sin(x)] = x cos (x) + sin(x).dxCOSIntegrating both sides indefinitely and using the fact that the integral of a sum is thesum of the integrals, we find thatdsin(x)]dxdxх cos(x) da +sin(x) dx.In this last equation, evaluate the indefinite integral on the left side:S (æ sin(x)]) dæ =-XCOSX + Sinx + CNow evaluate the indefinite integral on the right side:S sin(x) dæ-COSX + C(ii) Given the information calculated in (i), we can now determine thatSx cos(x) dxxsinx + coS X + C(iii) For which product of basic functions have you now found the antiderivative?Answer:COSX + xsinx Previously, we developed the Product Rule and studied how it is employed todifferentiate a product of two functions. In particular, recall that if f and g aredifferentiable functions of x, thend[f(x)•g(x)] = f(x)·g'(x) + g(x) · f'(x).da(a) For each of the following functions, use the Product Rule to find the function'sderivative. Notice the label of the derivative (e.g., the derivative of g(x) should belabeled g'(x)).(1) If g(x) = x sin(x), then g'(x) =XCOSx + sinx(ii) If h(x) = xe², then h'(x)xe^x + e^x(ii) If p(x) = x lIn(x), then p'(x)1+Inx(iv.) If q(æ) = x² cos(x), then q'(x)-x^2sinx + 2xcOsx(v.) If r(x) = e" sin(x), then r'(x)e^x (cosx + sinx)(b) Use your work in (a) to help you evaluate the following indefinite integrals. Usedifferentiation to check your work. (Don't forget the "+C".)(i)xe" + e" dx =xe^x +C(ii) / e"(sin(x) + cos(x)) dæ =e^xsinx+C(ii)2x cos(x) – a° sin(x) dæ =x^2cosx + C(iv) / æ cos(x) + sin(æ) dæ =xsinx + C(v)1+ In(x) dæ =xlnx + C(c) Observe that the examples in (b) work nicely because of the derivatives you wereasked to calculate in (a). Each integrand in (b) is precisely the result of differentiatingone of the products of basic functions found in (a). To see what happens when anintegrand is still a product but not necessarily the result of differentiating anelementary product, we consider how to evaluatex cos(x) dx.() First, observe thatd[æ sin(x)] = x cos(x)+ sin(æ).daIntegrating both sides indefinitely and using the fact that the integral of a sum is thesum of the integrals, we find thatSt in(a) dz = /a cos(a) de - / nin(e) dr.da =In this last equation, evaluate the indefinite integral on the left side:S (#(# sin(æ)]) dæ =-xcosx + sinx + CNow evaluate the indefinite integral on the right side:S sin(x) dx =-cosx + C(ii) Given the information calculated in (i), we can now determine thatSæ cos(a) dæ =xsinx + coS x + C(iii) For which product of basic functions have you now found the antiderivative?Answer:CoSx + xsinx

Answered: Image /qna-images/answer/31fb8be0-176c-412b-bf05-ef4f80d909b9.jpg

(c) Observe that the examples in (b) work nicely because of the derivatives you were asked to calculate in (a). Each integrand in (b) is precisely the result of differentiating one of the products of basic functions found in (a). To see what happens when an integrand is still a product but not necessarily the result of differentiating an elementary product, we consider how to evaluate x cos(x) dx. (i) First, observe that d -[x sin(x)] = x cos(x) + sin(x). dx Integrating both sides indefinitely and using the fact that the integral of a sum is the sum of the integrals, we find that d [x sin(x)] ) dx x cos(x) dx + sin(x) dx. dx In this last equation, evaluate the indefinite integral on the left side: S( sin(x)]) dæ = -XCOSX + Sinx + C Now evaluate the indefinite integral on the right side: