Question

and r(x) = 3x² If P(x)=x²-3x²-1/2 x + 10, Q(x) = ²x² - 1x ²-1/2 X- subtract R(x) from the sum of P(x) and Q(x). 6x + 1 then find p(x) + q(x) - r(x). 3 5 and R(x) - 34 X + the 4X​

Answer 1

Answer:Z

cos x dx = sin x + C

Z

sec2 x dx = tan x + C

Z

sec x tan x dx = sec x + C

Z

1

1 + x2

dx = arctan x + C

Z

1

√

1 − x2

dx = arcsin x + C

8.1 Substitution

Needless to say, most problems we encounter will not be so simple. Here’s a slightly more

complicated example: find

Z

2x cos(x

2

) dx.

This is not a “simple” derivative, but a little thought reveals that it must have come from

an application of the chain rule. Multiplied on the “outside” is 2x, which is the derivative

of the “inside” function x

2

. Checking:

d

dx sin(x

2

) = cos(x

2

)

d

dxx

2 = 2x cos(x

2

),

so Z

2x cos(x

2

) dx = sin(x

2

) + C.

Even when the chain rule has “produced” a certain derivative, it is not always easy to

see. Consider this problem:

Z

x

3

p

1 − x

2 dx.

There are two factors in this expression, x

3

and p

1 − x

2, but it is not apparent that the

chain rule is involved. Some clever rearrangement reveals that it is:

Z

x

3

p

1 − x

2 dx =

Z

(−2x)

−

1

2

(1 − (1 − x

2

))p

1 − x

2 dx.

This looks messy, but we do now have something that looks like the result of the chain

rule: the function 1 − x

2 has been substituted into −(1/2)(1 − x)

√

x, and the derivative

Step-by-step explanation:Z

cos x dx = sin x + C

Z

sec2 x dx = tan x + C

Z

sec x tan x dx = sec x + C

Z

1

1 + x2

dx = arctan x + C

Z

1

√

1 − x2

dx = arcsin x + C

8.1 Substitution

Needless to say, most problems we encounter will not be so simple. Here’s a slightly more

complicated example: find

Z

2x cos(x

2

) dx.

This is not a “simple” derivative, but a little thought reveals that it must have come from

an application of the chain rule. Multiplied on the “outside” is 2x, which is the derivative

of the “inside” function x

2

. Checking:

d

dx sin(x

2

) = cos(x

2

)

d

dxx

2 = 2x cos(x

2

),

so Z

2x cos(x

2

) dx = sin(x

2

) + C.

Even when the chain rule has “produced” a certain derivative, it is not always easy to

see. Consider this problem:

Z

x

3

p

1 − x

2 dx.

There are two factors in this expression, x

3

and p

1 − x

2, but it is not apparent that the

chain rule is involved. Some clever rearrangement reveals that it is:

Z

x

3

p

1 − x

2 dx =

Z

(−2x)

−

1

2

(1 − (1 − x

2

))p

1 − x

2 dx.

This looks messy, but we do now have something that looks like the result of the chain

rule: the function 1 − x

2 has been substituted into −(1/2)(1 − x)

√

x, and the derivative