integration sin^4x cos^3x dx
Answers
Answered by
14
Answer:
Step-by-step explanation:
Given:
To Find:
Solution:
We know that,
cos² x = (1 - sin²x)
Substituting the value,
Now let sin x = t
Differentiating on both sides,
cos x dx = dt
Substitute in the above integral,
Separating the integral,
Now give back the value of t,
This is the required integral.
Answered by
7
Written by FFdevansh
Your required answer mate
Answer:
\sf I= \dfrac{sin^5x}{5} -\dfrac{sin^7x}{7} +CI=
5
sin
5
x
−
7
sin
7
x
+C
Step-by-step explanation:
Given:
\sf sin^4x\:cos^3xsin
4
xcos
3
x
To Find:
\displaystyle \sf \int\limits {sin^4x\:cos^3x} \, dx∫sin
4
xcos
3
xdx
Solution:
\displaystyle \sf \int\limits {sin^4x\:cos^3x} \, dx∫sin
4
xcos
3
xdx
\displaystyle \sf \implies \int\limits {sin^4x\:cos^2x\:cos\:x} \, dx⟹∫sin
4
xcos
2
xcosxdx
We know that,
cos² x = (1 - sin²x)
Substituting the value,
\displaystyle \sf \implies \int\limits {sin^4x\:(1-sin^2x)\:cos\:x} \, dx⟹∫sin
4
x(1−sin
2
x)cosxdx
Now let sin x = t
Differentiating on both sides,
cos x dx = dt
Substitute in the above integral,
\displaystyle \sf \implies \int\limits {t^4(1-t^2)} \, dt⟹∫t
4
(1−t
2
)dt
\displaystyle \sf \implies \int\limits {t^4-t^6} \, dt⟹∫t
4
−t
6
dt
Separating the integral,
\displaystyle \sf \implies \int\limits {t^4} \, dt-\int\limits {t^6} \, dt⟹∫t
4
dt−∫t
6
dt
\sf \implies \dfrac{t^5}{5} -\dfrac{t^7}{7} +C⟹
5
t
5
−
7
t
7
+C
Now give back the value of t,
\sf \implies \dfrac{sin^5x}{5} -\dfrac{sin^7x}{7} +C⟹
5
sin
5
x
−
7
sin
7
x
+C
This is the required integral.
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