integral of sin²x-cos²x/sinxcosx
Answers
Answer:
log | secx / sinx | + c
Step-by-step explanation:
To find --->
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∫ ( Sin²x - Cos²x / Sinx Cosx ) dx
Solution--->
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∫ ( Sin² x - Cos²x / Sinx Cosx ) dx
=∫( Sin²x/Sinx Cosx)-(Cos²x/SinxCosx)dx
=∫ (Sinx / Cosx) dx - ∫ (Cosx / Sinx) dx
= ∫ tan x dx - ∫ Cot x dx
= log | sec x | - log | sinx | + c
We have a property of log
log m - log n = log m/n
Applying it here
=log | sec x/ sinx | + c
Additional information--->
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1) ∫ sec x dx = log (secx + tanx ) + c
2) ∫ cosec x dx = log tan x/2 + c
Answer:
Step-by-step explanation:
∫ ( Sin² x - Cos²x / Sinx Cosx ) dx
=∫( Sin²x/Sinx Cosx)-(Cos²x/SinxCosx)dx
=∫ (Sinx / Cosx) dx - ∫ (Cosx / Sinx) dx
= ∫ tan x dx - ∫ Cot x dx
= log | sec x | - log | sinx | + c
We have a property of log
log m - log n = log m/n
Applying it here