First slide
Introduction to integration
Question

 Evaluate 13sinx+cosxdx 

Easy
Solution

 Let 3=rsinθ and 1=rcosθ  Then r=(3)2+12=2 and tanθ=31 or θ=π3 

     13sinx+cosxdx    =1rsinθsinx+rcosθcosxdx    =1r1cos(xθ)dx=1rsec(xθ)dx 

    =1rlogtanπ4+x2θ2+c    =12logtanπ4+x2π6+c    =12logtanx2+π12+c

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