integral -1 to 1 log(1+sin x)/(1-sin x ) dx
Answers
Question :
Formula's used :
and ,
Solution :
Let
We know that :
Thus ,
Hence ,
Answer:
Formula's used :
\sf\green{We\:Know\:that}WeKnowthat
\sf\int\limits_{-a}^{a}f(x)\:dx=0,\:if\:f(-x)=-f(x)
−a
∫
a
f(x)dx=0,iff(−x)=−f(x)
and , \sf\int\limits_{-a}^{a}f(x)\:dx=2\int\limits_{0}^{a}f(x)\:dx,\:if\:f(-x)=f(x)
−a
∫
a
f(x)dx=2
0
∫
a
f(x)dx,iff(−x)=f(x)
Solution :
\sf\int\limits_{-1}^{1}\log(\frac{1+\sin\:x}{1-\sin\:x})dx
−1
∫
1
log(
1−sinx
1+sinx
)dx
Let \sf\:f(x)=\log(\frac{1+\sin\:x}{1-\sin\:x})f(x)=log(
1−sinx
1+sinx
)
\sf\:f(-x)=\log(\frac{1-\sin\:x}{1+\sin\:x})f(−x)=log(
1+sinx
1−sinx
)
We know that :
\sf-\log(\frac{a}{b})=\log(\frac{b}{a})−log(
b
a
)=log(
a
b
)
Thus ,
\sf\implies\:f(-x)=-\log(\frac{1+\sin\:x}{1-\sin\:x})⟹f(−x)=−log(
1−sinx
1+sinx
)
\sf\implies\:f(-x)=-f(x)⟹f(−x)=−f(x)
Hence ,
\tt\int\limits_{-1}^{1}\log(\frac{1+\sin\:x}{1-\sin\:x})dx=0
−1
∫
1
log(
1−sinx
1+sinx
)dx=0