Evaluate the following integral ∫_0^(π/2)〖e^x (sinx+cosx ) 〗 dx
I=∫_0^(π/2)〖e^x sinx dx〗+∫_0^(π/2)〖e^x cosx 〗 dxLet u=e^x dv=sinx dx du=e^x dx v=-cosx∴I_1=-├ e^x cosx ┤|_0^(π/2)+∫_0^(π/2)〖e^x cosx 〗 dx=e^(π/2)+I_2u=e^x dv=cosx dx du=e^x dx v=sinx∴I_1=e^(π/2)+├ e^x sinx ┤|_0^(π/2)-I_1=2e^(π/2)-I_1 ∴I_1=e^(π/2)Also:I_2=├ e^x sinx ┤|_0^(π/2)-∫〖e^x sinx 〗 dx=e^(π/2)-e^(π/2)=0∴I=e^(π/2)
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I=∫_0^(π/2)〖e^x sinx dx〗+∫_0^(π/2)〖e^x cosx 〗 dx
need an explanation for this answer? contact us directly to get an explanation for this answerLet u=e^x dv=sinx dx
du=e^x dx v=-cosx
∴I_1=-├ e^x cosx ┤|_0^(π/2)+∫_0^(π/2)〖e^x cosx 〗 dx=e^(π/2)+I_2
u=e^x dv=cosx dx
du=e^x dx v=sinx
∴I_1=e^(π/2)+├ e^x sinx ┤|_0^(π/2)-I_1=2e^(π/2)-I_1
∴I_1=e^(π/2)
Also:
I_2=├ e^x sinx ┤|_0^(π/2)-∫〖e^x sinx 〗 dx=e^(π/2)-e^(π/2)=0
∴I=e^(π/2)