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Solution

∫x sinx sec^3x dx =∫x sinx 1/(cos^3 x) dx =∫x tanx sec^2 x dx =x∫secx(secx tanx )dx-∫[secx(secx tanx )dx ]dx+C =x sec^2x/2-∫sec^2x/2 dx+C =x sec^2x/2-tanx/2+C

∫x sinx sec^3x dx =∫x sinx 1/(cos^3 x) dx =∫x tanx sec^2 x dx =x∫secx(secx tanx )dx-∫[secx(secx tanx )dx ]dx+C =x sec^2x/2-∫sec^2x/2 dx+C =x sec^2x/2-tanx/2+C

Solution

I=∫√(e^x-1 ) dx Let e^x-1=t^2⇒e^x dx=2t dt ⇒dx=2t/(t^2+1) dt ⇒I=∫t 2t/(t^2+1) dt=∫(2t^2)/(t^2+1) dt =∫(2(t^2+1)-2)/(t^2+1) dt=∫2dt-∫2dt/(t^2+1) =2t-2 tan^(-1)t+C =2√(e^x-1)-2 tan^(-1)√(e^x-1)+C

I=∫√(e^x-1 ) dx Let e^x-1=t^2⇒e^x dx=2t dt ⇒dx=2t/(t^2+1) dt ⇒I=∫t 2t/(t^2+1) dt=∫(2t^2)/(t^2+1) dt =∫(2(t^2+1)-2)/(t^2+1) dt=∫2dt-∫2dt/(t^2+1) =2t-2 tan^(-1)t+C =2√(e^x-1)-2 tan^(-1)√(e^x-1)+C

Solution

∫(e^x (x^2+1))/(x+1)^2 dx =∫(e^x (x^2-1+2))/(x+1)^2 dx =∫e^x [(x-1)/(x+1)+2/(x+1)^2 ]dx =∫e^x [f(x)+f^' (x)]dx,where f(x)=(x-1)/(x+1) and f^'(x)=2/(x+1)^2 =e^x ((x-1)/(x+1))+C

∫(e^x (x^2+1))/(x+1)^2 dx =∫(e^x (x^2-1+2))/(x+1)^2 dx =∫e^x [(x-1)/(x+1)+2/(x+1)^2 ]dx =∫e^x [f(x)+f^' (x)]dx,where f(x)=(x-1)/(x+1) and f^'(x)=2/(x+1)^2 =e^x ((x-1)/(x+1))+C

Solution

I=-e^(-x) log(e^x+1)+∫(e^(-x) e^x)/(e^x+1) dx = -e^(-x) log(e^x+1)+∫e^(-x)/(e^(-x)+1) dx = -e^(-x) log(e^x+1)-log(e^(-x)+1)+C = -e^(-x) log(e^x+1)-log(1+e^x )+x+C = -(e^(-x)+1) log(e^x+1)+x+C

I=-e^(-x) log(e^x+1)+∫(e^(-x) e^x)/(e^x+1) dx = -e^(-x) log(e^x+1)+∫e^(-x)/(e^(-x)+1) dx = -e^(-x) log(e^x+1)-log(e^(-x)+1)+C = -e^(-x) log(e^x+1)-log(1+e^x )+x+C = -(e^(-x)+1) log(e^x+1)+x+C

Solution

∫_a^bf(x)dx=[xf(x)]a^b-∫_a^bxf^'(x)dx (1) Now, put f(x)=t ∴ x=f^(-1 ) (t) and f^' (x)dx=dtand adjust the limits Therefore, ∫_a^bf(x)dx=[bf(b)-af(a) ]-∫_(f(a))^(f(b))f^(-1) (t)dt by (1) ∴∫_a^bf(x) +∫_(f(a))^(f(b))f^(-1) (x)dx=bf (b)-af(a)

∫_a^bf(x)dx=[xf(x)]a^b-∫_a^bxf^'(x)dx (1) Now, put f(x)=t ∴ x=f^(-1 ) (t) and f^' (x)dx=dtand adjust the limits Therefore, ∫_a^bf(x)dx=[bf(b)-af(a) ]-∫_(f(a))^(f(b))f^(-1) (t)dt by (1) ∴∫_a^bf(x) +∫_(f(a))^(f(b))f^(-1) (x)dx=bf (b)-af(a)

Solution

I_4,3=∫cos^4x sin3xdx Integrating by parts, we have I_4,3=-cos3x cos^4x /3-4/3 ∫cos^3x sinx cos3xdx But sinx cos3x=-sin2x+sin3x cosx , so I_4,3=-cosx cos^4x /3+4/3 ∫cos^3x sin2xdx-4/3 ∫cos^4x sin3xdx+C =-cos3x cos^4x/3+(4/3) I_3,2(-4/3) I_4,3+C Therefore, 7/3 I_4,3(-4/3) I_3,2=-cos3x cos^3x /3+C Or 7I_4,3-4I_3,2=-cos3x cos^4x+C

I_4,3=∫cos^4x sin3xdx Integrating by parts, we have I_4,3=-cos3x cos^4x /3-4/3 ∫cos^3x sinx cos3xdx But sinx cos3x=-sin2x+sin3x cosx , so I_4,3=-cosx cos^4x /3+4/3 ∫cos^3x sin2xdx-4/3 ∫cos^4x sin3xdx+C =-cos3x cos^4x/3+(4/3) I_3,2(-4/3) I_4,3+C Therefore, 7/3 I_4,3(-4/3) I_3,2=-cos3x cos^3x /3+C Or 7I_4,3-4I_3,2=-cos3x cos^4x+C