Using the technique of integration by parts evaluate the integral of the product of function [5+5] a) I=∫sin−1x dx b) I=∫x3ln⁡x dx

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Using the technique of integration by parts evaluate the integral of the product of function [5+5]  a) I=∫sin−1x dx  b) I=∫x3ln⁡x dx

ol3i4c5s4hr · Accepted Answer

Step 1  a) I=∫sin−1x dx By using integration of by points  I=∫1×sin−1x dx=sin−1x∫1dx−∫[ddx(sin−1x)∫1dx]dx  =sin−1x×x−∫[11−x2×x]dx  1) =xsin−1x−∫x1−x2dx  =I1 (say) let  I1=∫x1−x2dx let t=1−x2  dt=−2xdx  =∫1tdt−2  −dt2=xdx  =−12∫t−12dt  =−12[t1212]=−t12  I1=−1−x2 put in (1)  I=xsin−1x−(−1−x2)+c  Step 2  b) I=∫x3ln⁡x dx  By using integration of by parts  I=∫x3ln⁡x dx=ln⁡x∫x3dx−∫[ddxln⁡∫x3dx]dx

twineg4 · Accepted Answer

Step 1  Given: ∫x3×ln⁡(x)dx  Used formula:  ∫udv=uv−∫vdu  μ=ln⁡(x)⇒du=1xdx  dv=x3×dx⇒v=x44  So,  =x44×ln⁡(x)−∫x44×1xdx  =x44ln⁡(x)−14∫x3dx  =x44ln⁡(x)−14x44  =x44(ln⁡(x)−14)+C

nick1337 · Accepted Answer

Step 1 Use integration by parts, which states that: ∫udv=uv−∫vdu So, for ∫x3ln⁡xdx, let u=ln⁡x and dv=x3dx These imply that  du=1xdx and v=x44 (obtain these by differentiating u and integrating dv, respectively). Plugging these into the integration by parts formula, this yields: ∫x3ln⁡xdx=ln⁡x(x44)−∫(x44)(1x)dx =x4ln⁡x4−14∫x3dx =xln⁡x4−x416+C =x4(4ln⁡x−1)16+C

Using the technique of integration by parts evaluate the integral

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