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

Chaya Galloway

Chaya Galloway

Answered question

2021-10-18

Express 1(3t+1)(t+1) in partial fractions.
By using the substitution t=tanx , or otherwise, show that 0π4dx3+5sin2x=18ln2

Answer & Explanation

Yusuf Keller

Yusuf Keller

Skilled2021-10-19Added 90 answers

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)=321(3t+1)121(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,
sinx=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)

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