Question

If y=
1
y=sin x
cos y cos 2r
=
1
1
+===+
cos 2x cos 3x
cos 9x cos 10x
then
dy
A. 10 sec 10x
B. secʻx
sec
C. sec?x-10 sec 10x
D. 10 sec 10x-secx​

Answer 1

Answer:

y=sin2x−x+2sinx−log∣secx+tanx∣+C.

We have,

dx

dy

=

cosx

cos3x+cos2x

⇒

dx

dy

=

cosx

4cos

3

x−3cosx+2cos

2

x−1

⇒

dx

dy

=4cos

2

x−3+2cosx−secx

⇒

dx

dy

=2×2cos

2

x−2+2−3+2cosx−secx

⇒

dx

dy

=2cos2x+2−3+2cosx−secx

Integrating we get,

⇒y=sin2x−x+2sinx−log(secx+tanx)+C.

Answer 2

Answer:

may this help you

Step-by-step explanation:

y=sin2x−x+2sinx−log∣secx+tanx∣+C.

We have,

dx

dy

=

cosx

cos3x+cos2x

⇒

dx

dy

=

cosx

4cos

3

x−3cosx+2cos

2

x−1

⇒

dx

dy

=4cos

2

x−3+2cosx−secx

⇒

dx

dy

=2×2cos

2

x−2+2−3+2cosx−secx

⇒

dx

dy

=2cos2x+2−3+2cosx−secx

Integrating we get,

⇒y=sin2x−x+2sinx−log(secx+tanx)+C.