Use a table of integrals to evaluate the definite integral. ∫0π2xsin⁡2xdx

Question

Use a table of integrals to evaluate the definite integral.  ∫0π2xsin⁡2xdx

Philip Williams · Accepted Answer

Step 1Given: I=∫0π2xsin⁡(2x)dxfor evaluating given integral, we use integral by parts theoremaccording to this theorem∫(f′(x))g(x)dx=g(x)∫f′(x)dx−∫[(g′(x))∫f′(x)dx]dx+cStep 2so,∫0π2xsin⁡(2x)=[x∫sin⁡2xdx−∫[1∫sin⁡2xdx]dx]0π2(∵∫sin⁡kxdx=−cos⁡kxk+c)=[x(−cos⁡2x2)−∫(−cos⁡2x2)dx]0π2(∵∫cos⁡kxdx=sin⁡kxk+c)=[−xcos⁡2x2+12sin⁡2x2]0π2=[−2xcos⁡2x+sin⁡2x4]0π2=14[(−2(π2)cos⁡(π)+sin⁡(π))−(−2(0)cos⁡(0)+sin⁡(0))]=14[−π(−1)+0+0−0]=14(π)=π4hence, given integral is equal to

veiga34 · Accepted Answer

∫xsin⁡(2x)dx  =−xcos⁡(2x)2−∫−cos⁡(2x)2dx  ∫−cos⁡(2x)2dx  =−14∫cos⁡(u)du  ∫cos⁡(u)du  =sin⁡(u)  −14∫cos⁡(u)du  =−sin⁡(u)4  =−sin⁡(2x)4  −xcos⁡(2x)2−∫−cos⁡(2x)2dx  =sin⁡(2x)4−xcos⁡(2x)2  ∫xsin⁡(2x)dx  =sin⁡(2x)4−xcos⁡(2x)2+C  =sin⁡(2x)−2xcos⁡(2x)4+C

karton · Accepted Answer

∫0π2x×sin⁡(2x)dx Evaluate the indefinite integral x×sin⁡(2x)dxUse the partial integrationx×(−cos⁡(2x)2)−∫−cos⁡(2x)2dxx×(−cos⁡(2x)2)−1×(−12)×∫cos⁡(2x)dxx×(−cos⁡(2x)2)+12×∫cos⁡(t)2dtx×(−cos⁡(2x)2)+12×12×∫cos⁡(t)dtx×(−cos⁡(2x)2)+14×sin⁡(t)x×(−cos⁡(2x)2)+14×sin⁡(2x)−x×cos⁡(2x)2+sin⁡(2x)4(−x×cos⁡(2x)2+sin⁡(2x)4)|0π2−π2×cos⁡(2×π2)2+sin⁡(2×π2)4−(−0cos⁡(2×0)2+sin⁡(2×0)4) Simplify Solution π4

Use a table of integrals to evaluate the definite integral. \int_{0}^{\frac{\pi}{2}}x\sin

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