Express 1(3t+1)(t+1) in partial fractions. By using the substitution t=tan⁡x , or otherwise, show that ∫0π4dx3+5sin⁡2x=18ln⁡2

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Express 1(3t+1)(t+1) in partial fractions.  By using the substitution t=tan⁡x , or otherwise, show that ∫0π4dx3+5sin⁡2x=18ln⁡2

Yusuf Keller · Accepted Answer

Step 1 First find the partial fraction of the given term. 1(3t+1)(t+1)=A(3t+1)+Bt+1 1(3t+1)(t+1)=A(t+1)+B(3t+1)(3t+1)(t+1) Now taking the numerators of both sides. 1=A(t+1)+B(3t+1)   ……. (1) Put t=−1 in equation (1). 1=A(−1+1)+B(3(−1)+1) 1=A(0)+B(−2) B=−12 Put t=0 and the value of B in equation (1). 1=A(0+1)+B(3(0)+1) 1=A+(−12) A=1+12 A=32 Hence the partial fraction the given term is. 1(3t+1)(t+1)=32⋅1(3t+1)−12⋅1(t+1)      ………… (2) Step 2  NOTE- There is a mistake in the question, the terms on the R.H.S. should be 18ln⁡(3) To prove. ∫0π4dx3+5sin⁡(2x)=18ln⁡(3) Recall the formula, sin⁡x=2tan⁡(x2)1+tan2(x2) So, sin⁡(2x)=2tan⁡(2x2)1+tan2(2x2)=2tan⁡(x)1+tan2(x) Now on using this formula the integral becomes. I=∫0π4dx3+5sin⁡(2x)

Express \frac{1}{(3t+1)(t+1)} in partial fractions. By using the subst

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